IIT-JEE 2011 - Career Point
IIT-JEE 2011 - Career Point IIT-JEE 2011 - Career Point
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- Page 5 and 6: Volume-5 Issue-11 May, 2010 (Monthl
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No success is possible unless you believe that you can succeed.<br />
Volume - 5 Issue - 11<br />
May, 2010 (Monthly Magazine)<br />
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Editor : Pramod Maheshwari<br />
Dear Students,<br />
Find a mentor who can be your role model and your friend !<br />
A mentor is someone you admire and under whom you can study.<br />
Throughout history, the mentor-protege relationship has proven quite<br />
fruituful. Socrates was one of the early mentors. Plato and Aristotle studied<br />
under him and later emerged as great philosophers in their own right.<br />
Some basic rules to know mentors :<br />
• The best mentors are successful people in their own field. Their behaviors<br />
are directly translatable to your life and will have more meaning to you.<br />
• Be suspicious of any mentors who seek to make you dependent on them.<br />
It is better to have them teach you how to fish than to have them catch<br />
the fish for you. That way, you will remain in control.<br />
• Turn your mentors into role models by examining their positive traits.<br />
Write down their virtues. without identifying to whom they belong.<br />
When you are with these mentors, look for even more behavior that<br />
reflect their success. Use these virtues as guidelines for achieving<br />
excellence in your field.<br />
Be cautious while searching for a mentor :<br />
• Select people to be your mentors who have the highest ethical standards<br />
and a genuine willingness to help others.<br />
• Choose mentors who have and will share superb personal development<br />
habits with you and will encourage you to follow suit.<br />
• Incorporate activities into your mentor relationship that will enable your<br />
mentor to introduce you to people of influence or helpfulness.<br />
• Insist that your mentor be diligent about monitoring your progress with<br />
accountability functions.<br />
• Encourage your mentor to make you an independent, competent, fully<br />
functioning, productive individual. (In other words, give them full<br />
permission to be brutally honest about what you need to change.)<br />
Getting benefited from a role-mode :<br />
Acquiring good habits from others will accelerate you towards achieving your<br />
goals. Ask yourself these questions to get the most out of your role<br />
model/mentors :<br />
• What would they do in my situation?<br />
• What do they do every day to encourage growth and to move closer to a<br />
goal ?<br />
• How do they think in general ? in specific situations ?<br />
• Do they have other facts of life in balance ? What effect does that have on<br />
their well-being ?<br />
• How do their traits apply to me ?<br />
• Which traits are worth working on first ? Later ?<br />
A final word : Under the right circumstances mentors make excellent role<br />
models. The one-to-one setting is highly conducive to learning as well as to<br />
friendship. But the same cautions hold true here as for any role model. It is<br />
better to adapt their philosophies to your life than to adopt them.<br />
Presenting forever positive ideas to your success.<br />
Yours truly<br />
Pramod Maheshwari,<br />
B.Tech., <strong>IIT</strong> Delhi<br />
Editorial<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 1 MAY 2010
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 2 MAY 2010
Volume-5 Issue-11<br />
May, 2010 (Monthly Magazine)<br />
NEXT MONTHS ATTRACTIONS<br />
Key Concepts & Problem Solving strategy for <strong>IIT</strong>-<strong>JEE</strong>.<br />
INDEX<br />
CONTENTS<br />
Regulars ..........<br />
PAGE<br />
Know <strong>IIT</strong>-<strong>JEE</strong> With 15 Best Questions of <strong>IIT</strong>-<strong>JEE</strong><br />
Challenging Problems in Physics, Chemistry & Maths<br />
Much more <strong>IIT</strong>-<strong>JEE</strong> News.<br />
Xtra Edge Test Series for <strong>JEE</strong> – <strong>2011</strong> & 2012<br />
AIEEE-2010 Examination Paper<br />
NEWS ARTICLE 4<br />
<strong>IIT</strong>-Develops technology to produce stealth aircraft<br />
Urine-processing technologies yield rich cash flow potential<br />
<strong>IIT</strong>ian ON THE PATH OF SUCCESS 8<br />
Mr. Sujal Patel<br />
KNOW <strong>IIT</strong>-<strong>JEE</strong> 10<br />
Previous <strong>IIT</strong>-<strong>JEE</strong> Question<br />
Study Time........<br />
DYNAMIC PHYSICS 17<br />
S<br />
Success Tips for the Months<br />
• "All of us are born for a reason, but all of<br />
us don't discover why. Success in life has<br />
nothing to do with what you gain in life or<br />
accomplish for yourself. It's what you do<br />
for others."<br />
• "Don't confuse fame with success.<br />
Madonna is one; Helen Keller is the<br />
other."<br />
• "Success is not the result of spontaneous<br />
combustion. You must first set yourself on<br />
fire."<br />
• "Success does not consist in never making<br />
mistakes but in never making the same one<br />
a second time."<br />
• "A strong, positive self-image is the best<br />
possible preparation for success."<br />
• "Failure is success if we learn from it."<br />
• "The first step toward success is taken<br />
when you refuse to be a captive of the<br />
environment in which you first find<br />
yourself."<br />
8-Challenging Problems [Set# 1]<br />
Students’ Forum<br />
Physics Fundamentals<br />
Electrostatics-I<br />
1- D Motion, Projectile Motion<br />
CATALYSE CHEMISTRY 33<br />
XTRAEDGE TEST SERIES 49<br />
Class XII – <strong>IIT</strong>-<strong>JEE</strong> <strong>2011</strong> Paper<br />
Class XI – <strong>IIT</strong>-<strong>JEE</strong> 2012 Paper<br />
Key Concept<br />
Gaseous State & Real Gases<br />
General organic Chemistry<br />
Understanding : Physical Chemistry<br />
DICEY MATHS 41<br />
Mathematical Challenges<br />
Students’ Forum<br />
Key Concept<br />
Complex Number<br />
Matrices & Determinants<br />
Test Time ..........<br />
<strong>IIT</strong> - 2010 Examination Paper with Solution 66<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 3 MAY 2010
<strong>IIT</strong> develops technology<br />
to produce stealth<br />
aircraft<br />
Materials scientists at the Indian<br />
Institute of Technology in<br />
Roorkee (<strong>IIT</strong>-R) have developed<br />
microwave absorbing nanocomposite<br />
coatings that could make<br />
aircraft almost invisible to radar.<br />
The technology for building<br />
invisible, or stealth aircraft, is a<br />
closely guarded secret of<br />
developed countries and a handful<br />
of laboratories in India are doing<br />
research in this area.<br />
Radars that emit pulses of<br />
microwave radiation identify flying<br />
aircraft by detecting the radiation<br />
reflected by the aircraft’s metallic<br />
body. The nanocomposite coatings<br />
developed by Rahul Sharma, R.C.<br />
Agarwala and Vijaya Agarwala at<br />
<strong>IIT</strong>-R absorb most of the incident<br />
radiation and reflect very little.<br />
Sharma, who revealed his team’s<br />
work at an international<br />
nanomaterials conference held<br />
recently at the Indian Institute of<br />
Science in Bangalore, believes<br />
their nano-product is a significant<br />
step in developing a technology to<br />
enable aircraft escape radar<br />
surveillance and protect its<br />
equipment from electronic<br />
“jamming”.Nanoparticles - so<br />
called because of their very small<br />
size - are known to exhibit unique<br />
physical and chemical properties.<br />
The <strong>IIT</strong> team found that crystals of<br />
“barium hexaferrite” with particle<br />
size of 10-15 nanometres have the<br />
ability to absorb microwaves.<br />
(Human hair, for comparison, is<br />
100,000 nanometres thick). They<br />
developed special processes for<br />
synthesizing the nanopowder and<br />
formulating it as a coating.<br />
Sharma said that the<br />
nanocomposite coating on the<br />
aluminium sheet absorbed 89<br />
percent of incident microwaves at<br />
15 giga hertz - the frequency<br />
normally used by radars —<br />
reflecting only 11 percent. A<br />
stealth aircraft should ideally<br />
absorb all the incident radiation<br />
and reflect nothing.<br />
Urine-processing<br />
technologies yield rich<br />
cash flow potential<br />
The stink is out of urine, literally<br />
and metaphorically, with a growing<br />
number of researchers spotting<br />
commercial and ecological value in<br />
a liquid most people consider<br />
waste.<br />
The Indian Institute of Technology<br />
(<strong>IIT</strong>) Delhi, for instance, is working<br />
to harvest this human waste and<br />
convert it into fertiliser. The Delhi<br />
government is willing to consider<br />
a revenue-share commercial<br />
venture selling the phosphates and<br />
nitrates in urine.<br />
On the outskirts of Delhi, a littleknown<br />
non-government organisation,<br />
Fountain for Development<br />
Research and Action, is laying the<br />
ground for the first urine bank. It<br />
has diverted urine from two<br />
schools, where it has installed<br />
odour-free urinals, into a tank and<br />
transferred the run-off to a village<br />
nearby for use as fertiliser.<br />
Director Madhab Nayak says the<br />
foundation is working towards<br />
making farmers aware of its<br />
potential as replacement for<br />
expensive urea.<br />
"There is no such thing as waste,"<br />
says Vijayaraghavan M Chariar,<br />
assistant professor at the Centre<br />
for Rural Development and<br />
Technology at <strong>IIT</strong>. "Urine consists<br />
of a lot of inorganic salts, which<br />
produce gases only when mixed<br />
with water. It is, in fact, pure<br />
fertiliser," he added.<br />
<strong>IIT</strong> has come up with a cheap,<br />
odour-free, urinal which it has<br />
successfully tested on campus. The<br />
odour-free urinal combines<br />
technology with simple science to<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 4 MAY 2010
translate into a significant watersaving<br />
initiative (urine smells only<br />
when mixed with water, which<br />
this technology eliminates).<br />
Urine is collected through a tank<br />
placed underground, harvested<br />
and used as liquid fertiliser two to<br />
three metres below the ground on<br />
a five-acre field on campus, said<br />
Chariar, who can talk animatedly<br />
about this human waste and how<br />
its poor treatment alone has led<br />
to sanitation problems.<br />
The public urinal at <strong>IIT</strong> uses a<br />
simple technology, called Zerodor,<br />
developed by Chariar, that fits<br />
into the waste coupler in the pan<br />
and diverts the urine through a<br />
drain where it is collected and<br />
harvested. The idea is not to<br />
allow it to mix with water at any<br />
stage.<br />
Chariar has already transferred<br />
this technology to Good Yield<br />
Environmental Technologies, a<br />
Kolkata firm, and filed for a patent.<br />
Chariar claims that Zerodor is a<br />
low cost product and would need<br />
replacement in only about two<br />
years.<br />
Meanwhile, the Delhi government,<br />
which has already installed 200<br />
such odourless urinals in different<br />
parts of the city, uses a different<br />
and perhaps more expensive<br />
technology. Amiya Chandra,<br />
deputy commissioner of the city’s<br />
municipal corporation, says,<br />
"Other than problems of<br />
vandalism, these urinals are<br />
working perfectly."<br />
In preparation for the<br />
Commonwealth Games, the Delhi<br />
government is planning to install<br />
1,000 such urinals at a nominal<br />
cost of Rs 3 lakh.<br />
Chariar is already working on the<br />
second phase of his project, which<br />
was initiated by Unicef and<br />
Stockholm Environment Institute,<br />
for setting up a small reactor to<br />
extract nitrates and phosphates<br />
from urine. "This could become a<br />
micro-enterprise from the urinal,"<br />
says Chariar.<br />
The Delhi government is also<br />
looking at installing Chariar’s<br />
technology at a few parks in the<br />
city, while harvesting urine in<br />
those places.<br />
Chariar has even designed similar<br />
urinals for women. "We have filed<br />
for trademark registration and we<br />
are in discussion with companies<br />
for marketing it," he says. With a<br />
little more investment, he says, a<br />
hydrophobic coating on pans<br />
could make it water resistant and<br />
completely drain the urine, leaving<br />
no room for any oxidisation,<br />
which can also cause odour.<br />
In the developed world,<br />
communities have been quick to<br />
realise the huge economic<br />
potential of urine. "Communities<br />
in Germany are exporting urine to<br />
neighbouring countriesthat are<br />
using it on their farms, says<br />
Chariar, explaining how it could<br />
be diverted for use as a nutrient<br />
by a simple plumbing.<br />
The urine tank could deliver the<br />
liquid nutrient directly to plants<br />
about two to three metres below<br />
the soil, he says.<br />
The Centre for Banana Research<br />
in Trichy is already using it for<br />
banana plantations and the<br />
University of agriculture Sciences,<br />
Bangalore, too is looking at its<br />
varied uses.<br />
<strong>IIT</strong> student produces<br />
electricity from waste<br />
water<br />
Kolkata: Waste water<br />
management is a big issue world<br />
wide and specially in India where<br />
there is acute shortage of the<br />
precious resource in many places<br />
but a 23-year-old student of <strong>IIT</strong><br />
Kharagpur claims he has found a<br />
solution.<br />
Apart from finding solutions<br />
management of waste water he<br />
has also demonstrated producing<br />
electricity from it, which could go<br />
a long way in protecting the<br />
earth's resources.<br />
A look at reservoirs used<br />
for water supply<br />
Manoj Mandelia, who is pursuing<br />
integrated MTech at <strong>IIT</strong><br />
Kharagpur, there was no policy in<br />
the country which examined<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 5 MAY 2010
waste as part of a cycle of<br />
production-consumption-ecovery.<br />
"Waste management still<br />
constituted a linear system of<br />
collection and disposal which<br />
creates health and environmental<br />
hazards," he said.<br />
"I developed a product which uses<br />
the concept of microbial fuel cell<br />
(MFC is a bio-electrochemical<br />
system that drives a current by<br />
mimicking bacterial interactions<br />
found in nature), which could not<br />
only treat waste water but also<br />
produce electricity in the<br />
process," explains Mandelia who<br />
heads a team of five people in the<br />
project.<br />
The Water Diviner<br />
The project, named LOCUS which<br />
stands for Localised Operation of<br />
Bio-cells Using Sewage, can<br />
achieve chemical oxygen demand<br />
(COD) reduction levels in waste<br />
water to about 60-80 per cent.<br />
<strong>IIT</strong> Kharagpur Calls for<br />
Nominations for the<br />
Nina Saxena Excellence<br />
in Technology Award<br />
<strong>IIT</strong> Kharagpur announces the<br />
fourth edition of Nina Saxena<br />
Excellence in Technology Award<br />
to invite entries of this year in<br />
areas of technical innovations.<br />
Nominations for entries to the<br />
award are open until April 30,<br />
2010. Entries can be submitted<br />
either by post to the Director, <strong>IIT</strong><br />
Kharagpur, West Bengal,PIN -<br />
721302, India or by email at<br />
director@iitkgp.ernet.in. The<br />
nomination form is also available<br />
on the official award website<br />
The objective behind the award is<br />
to commemorate the spirit and<br />
drive of Dr. Nina Saxena, who<br />
personified technical excellence.<br />
The fourth edition will continue<br />
the tradition of rewarding<br />
pioneering innovations for<br />
betterment of society.<br />
A distinguished committee is being<br />
formed by <strong>IIT</strong> Kharagpur to<br />
adjudge the nominations for this<br />
award. The award committee is<br />
chaired by Director of <strong>IIT</strong><br />
Kharagpur and is comprised of<br />
Deans and selected faculty<br />
members of <strong>IIT</strong> Kharagpur and<br />
well known alumnus, based in<br />
India and in the US.<br />
<strong>IIT</strong>’s New On-Campus<br />
Wind Turbine to<br />
Support Green Jobs,<br />
Research, and Education<br />
The consortium members will<br />
research the wind energy<br />
challenges identified in the U.S.<br />
Department of Energy's "20%<br />
Wind Energy by 2030" report,<br />
including wind technology, grid<br />
system integration, and workforce<br />
challenges. The consortium's plan<br />
relies on <strong>IIT</strong> experts in electrical<br />
and computer engineering;<br />
mechanical, materials, and<br />
aerospace<br />
engineering;<br />
architecture; business; and<br />
members of the Wanger Institute<br />
for Sustainable Energy Research to<br />
tackle these challenges.<br />
Many of the university's<br />
departments and research centers<br />
will also work together to offer<br />
wind energy courses addressing<br />
the technical, operational, social,<br />
and environmental aspects of wind<br />
energy in consultation with<br />
industry. To ensure student<br />
involvement in the project,<br />
fellowships will be offered annually<br />
to undergraduate and graduate<br />
students in wind energy<br />
engineering fields of study. Faculty<br />
and students from international<br />
university consortium members<br />
will also be invited to <strong>IIT</strong> to attend<br />
workshops and to share ideas with<br />
their American counterparts.<br />
The wind energy consortium will<br />
work with small wind turbine<br />
manufacturer Viryd Technologies<br />
to procure and install an 8KW<br />
Viryd wind turbine on <strong>IIT</strong>'s Main<br />
Campus, and to deliver a second<br />
turbine to one of <strong>IIT</strong>'s engineering<br />
laboratories to perform turbine<br />
reliability studies. The consortium<br />
will also work with wind energy<br />
developer Invenergy to install a<br />
1.5MW GE wind turbine adjacent<br />
to a wind farm in Marseilles, Ill.<br />
The close proximity of <strong>IIT</strong>'s<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 6 MAY 2010
Marseilles turbine to an existing<br />
wind farm provides an ideal<br />
opportunity to study turbine-toturbine<br />
wake interaction, wind<br />
farm interaction, and wind energy<br />
efficiencies in addition to turbine<br />
reliability studies.<br />
Hyderabad boy tops in<br />
GATE 2010<br />
HYDERABAD : Malladi<br />
Harikrishna, a final year computer<br />
science engineering student from<br />
the city, has topped the national<br />
level Graduate Aptitude Test in<br />
Engineering (GATE-2010). He<br />
achieved the feat in his first<br />
attempt, scoring 99.99 percentile<br />
by scoring 83.55 per cent in<br />
GATE.<br />
The results were announced on<br />
March 15 by <strong>IIT</strong>-Guwahati, which<br />
conducted<br />
GATE this<br />
year. Mallad,<br />
who topped<br />
the nationallevel<br />
Graduate Aptitude Test in<br />
Engineering (GATE-2010), said,<br />
“My aim is to join ME at the Indian<br />
Institute of Science, Bengaluru. I<br />
want to become a scientist,”<br />
Harikrishna, 21, said.<br />
Students who clear GATE are<br />
eligible for admission to masters’<br />
degree courses in engineering,<br />
technology, architecture,<br />
pharmacy, science in premier<br />
institutes like <strong>IIT</strong>s and NITs.<br />
Many students from the state<br />
made it to the top-100.<br />
Srujana (JNTU, Kukatpally) ranked<br />
22, Karthik Nagarjuna 44,<br />
Mufaquam Ali 51 and Pavan<br />
Kishore got the 102nd rank in<br />
ECE stream.<br />
Srinivas Reddy got the 68th rank<br />
in EEE, and V. Suryanarayana 31 in<br />
Mechanical.<br />
GATE 2010 score is valid for two<br />
years from the date of<br />
announcement of the results,<br />
according to the details published<br />
on GATE website<br />
Terminated <strong>IIT</strong> students<br />
seek intervention of<br />
Prez,PM<br />
Terminated for "bad<br />
performance", 38 students of <strong>IIT</strong><br />
Kanpur have taken the issue to<br />
President Pratibha Patil and Prime<br />
Minister Manmohan Singh with a<br />
plea that the institute reconsider<br />
the decision.<br />
The students have written to<br />
Prime Minister and the PMO has<br />
forwarded the matter to the HRD<br />
Ministry for "appropriate action".<br />
The Ministry has again forwarded<br />
it to the institute for action, a<br />
ministry official said.<br />
The 38 students, including 24 from<br />
under-graduate and 14 postgraduate<br />
levels, were denied<br />
admission into fresh semester in<br />
January this year for "bad<br />
academic record"<br />
.Prof V N Pal, who is an alumni of<br />
the institute, has taken the<br />
students' issue to President Patil,<br />
who is the Visitor of the institute.<br />
Prof Pal met the President on<br />
April one at Rashtrapati Bhawan<br />
and discussed the issue of<br />
termination of students of <strong>IIT</strong><br />
Kanpur at length.<br />
He explained the socio-economic<br />
condition of these students.<br />
Vidya Balan to address<br />
<strong>IIT</strong><br />
She has been invited by one of the<br />
Indian Institutes of<br />
Technology to be a keynote<br />
speaker at an upcoming seminar.<br />
In an interview to a leading daily,<br />
she confirmed her invitation from<br />
one of the <strong>IIT</strong>s. She said, “I have<br />
been approached by one of the<br />
<strong>IIT</strong>s for being a keynote speaker at<br />
a seminar they are holding. The<br />
topic is ‘The changing face of the<br />
Indian heroine’. I am excited about<br />
it, but am figuring out dates and<br />
prior commitments at the<br />
moment. But, I hope this comes<br />
through.”<br />
A MA in sociology, Vidya Balan<br />
believes her education has<br />
groomed her and given her the<br />
confidence to address seminars at<br />
these prestigious universities.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 7 MAY 2010
Success Story<br />
This article contains story of a person who get succeed after graduation from different <strong>IIT</strong>'s<br />
MR. SUJAL PATEL<br />
To his competitors, Sujal Patel is now a name to reckon<br />
with. His company Isilon Systems, in the clustered storage<br />
space, has not only earned its position as the fastest<br />
growing technology company in North America, but in<br />
the five years of selling its products, Patel has transformed<br />
the company from zero sales, into a company with a $100<br />
million run rate, $80 million in cash, and no debt.<br />
“Slightly better is not a good term,” says Patel, the CEO<br />
of Isilon Systems and a pioneer in the clustered storage<br />
space. “For a technology to be adopted in an existing<br />
market, you really need to have a technology that is<br />
substantially better — 10 (times) better than what is in<br />
the marketplace today,” adds Patel. “It has to be so much<br />
better that it is overwhelming for people who buy that<br />
product and service.” When Patel says so, he is not being<br />
merely theoretical. The unsatiated desire to bring out the<br />
best led this 35 year old entrepreneur to steer his<br />
company successfully, in a market space, which was<br />
crowded by biggies like EMC and NetApp; not to mention<br />
the 250 odd startups in the storage space.<br />
Patel founded Isilon at the age of 26. Prior to this, Patel<br />
spent his initial career days at Real Networks. As an<br />
engineer, Patel used to solve some of Real Networks’<br />
most complex back-end operational challenges. That<br />
experience gave him the insight for a new type of solution,<br />
a type of virtualized storage optimized for media. His<br />
experience gave him the insight to a real customer need,<br />
and his deep technical knowledge gave him the ability to<br />
spot a solution.<br />
I did not want to wait 10-15 years, treading cautiously at<br />
every step before taking the plunge. So when I got the<br />
chance to found Isilon, I jumped at the opportunity,”<br />
beams Patel.<br />
That’s also a reason which led venture capitalists to take<br />
this 26 year old lad seriously. A few months after Patel<br />
founded Isilon in 2001, the NASDAQ came down<br />
crashing, bringing the ‘Dot Com Burst’. The venture<br />
capital market was in disarray. With the existing<br />
companies dropping their revenues, there was not much<br />
hope for new companies to find potential investment. To<br />
make matters worse, there were about 250 other<br />
startups in the storage space. Patel was undeterred. Sure<br />
about his ideas, he approached close to 40 venture<br />
capitalists, and with perseverance, eventually he managed<br />
to gain their confidence. Five months after the company’s<br />
beginning, Patel had managed to raise $8.4 million venture<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 8 MAY 2010
capital in what was one of the toughest markets to raise<br />
money.<br />
But all was not rosy yet. The road ahead proved to be<br />
more challenging than raising the funds. The next three<br />
years were spent in building the products. Developing the<br />
company’s IP not only took a tremendous amount of<br />
money, but also ate in to the time to get into the market.<br />
By the time the product was ready to debut in<br />
the markets in 2003, the company’s debt was near<br />
$20 million.<br />
Cruising over Obstacles<br />
Not only confident and unshakable, Patel is also a man of<br />
clear vision, with no illusions about his capabilities. At 26,<br />
as the Founder of the company, he served as the CEO for<br />
the initial three years. But he knew that, if he wanted to<br />
make Isilon the next big thing in storage space, he needed<br />
someone who knew the dynamics of running a large<br />
organization. He promptly stepped down and appointed<br />
as a new CEO, who knew what it takes to grow into the<br />
larger league. “I floated the company but knew my<br />
limitations in the business front. We had three products<br />
ready to debut in the market and if we wanted then to<br />
succeed, we needed an expert who knew the right strings<br />
to pull,” says Patel. His focus and timely decisions were<br />
fruitful in the subsequent years when Isilon was on a<br />
dream run, literally growing at 200 percent year-on-year.<br />
By the time the company went public, it was a $60 million<br />
company.<br />
“My goal was to see Isilon become a $100 million<br />
company by 2007 and become a player to be reckoned in<br />
the $4 billion global storage market for its technology.<br />
And we were still $40 million short. I knew that despite<br />
the economic instability, the company I had founded had<br />
great potential if one could maximize it,” says Patel.<br />
To start with, he re-structured the entire organization,<br />
replacing every executive in the management, including<br />
CTO, CFO, Head of engineering, Head of operations and<br />
others. Such an action is quite unheard of, especially<br />
immediately after a company had gone public.<br />
Next, Patel began revamping the business strategy by<br />
reaching out to the broader enterprise segments. It was<br />
not easy. The segment he wanted to target comprised of<br />
Fortune 50 companies who did not have much<br />
expectation from a startup like Isilon. With the<br />
organizational re-structuring, he completely overhauled<br />
the company’s services and products to meet the<br />
expectations of the large enterprise customers.<br />
Eventually, more and more Fortune 50 companies began<br />
to adopt Isilon’s products.<br />
Finally he decided to increase the company’s focus on<br />
R&D. Innovation has always been an essential part of his<br />
life. A very innovative and inquisitive person, Patel is<br />
known to get at least six ideas every day. Even as a<br />
kid he was known for asking around 500 questions<br />
about everything under the Sun. Thus, Patel diverted<br />
about 25 percent of the company’s revenue<br />
towards R&D and fostered innovation within the<br />
organization.<br />
So what drives this confident and zealous man? It is the<br />
self-belief, passion and the creative way of looking at<br />
problems and coming up with solutions, says Patel. Even<br />
during his college days, while working on the Internet, he<br />
looked for opportunities to think about new ways, solve<br />
problems, or to bring innovative techniques/technologies<br />
to the market place.<br />
To run a business successfully, it is also important that<br />
one should be honest with oneself, the team and the<br />
stakeholders. The foremost thing that Patel did after<br />
taking over the company was to communicate with<br />
customers and partners and update them about the<br />
happenings within the organization, reinstating their faith<br />
in him and the company. “I went and talked to each of our<br />
investors and customers, telling what we planned<br />
and how much earnings we expected through the<br />
quarters. I believe that apart from your technology<br />
offerings, one reason companies want to do business<br />
with you is the goodwill you develop in tough times,”<br />
says Patel.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 9 MAY 2010
KNOW <strong>IIT</strong>-<strong>JEE</strong><br />
By Previous Exam Questions<br />
PHYSICS<br />
1. In Searle's experiment, which is used to find Young's<br />
Modulus of elasticity, the diameter of experimental<br />
wire is D = 0.05 cm (measured by a scale of least<br />
count 0.001 cm) and length is L = 100 cm (measured<br />
by a scale of least count 0.1 cm). A weight of 50 N<br />
causes an extension of X = 0.125 cm (measured by a<br />
micrometer of least count 0.001 cm). Find maximum<br />
possible error in the values of Young's modulus.<br />
Screw gauge and meter scale are free from error.<br />
[<strong>IIT</strong>-2004]<br />
Sol. Maximum percentage error in Y is given by<br />
W L<br />
Y = ×<br />
2<br />
πD<br />
X<br />
4<br />
⎛ ∆Y ⎞<br />
⎜ ⎟⎠<br />
⎝ Y<br />
max.<br />
⎛ ∆D<br />
⎞<br />
= 2 ⎜ ⎟<br />
⎝ D ⎠<br />
∆ x + x<br />
∆L<br />
+ L<br />
⎛ 0.001⎞<br />
⎛ 0.001 ⎞ ⎛ 0.1 ⎞<br />
= 2 ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 0.0489<br />
⎝ 0.05 ⎠ ⎝ 0.125 ⎠ ⎝110<br />
⎠<br />
So maximum percentage error = 4.89%.<br />
2. Particles P and Q of mass 20 gm and 40 gm<br />
respectively are simultaneously projected from points<br />
A and B on the ground. The initial velocities of P and<br />
Q make 45º and 135º angles respectively with the<br />
horizontal AB as shown in the figure. Each particle<br />
has an initial speed of 49 m/s. The separation AB is<br />
245 m. [<strong>IIT</strong>-1982]<br />
P<br />
Q<br />
45º<br />
135º<br />
A<br />
B<br />
Both particle travel in the same vertical plane and<br />
undergo a collision After the collision, P retraces its<br />
path, Determine the position of Q when it hits the<br />
ground. How much time after the collision does the<br />
particle Q take to reach the ground? Take g = 9.8<br />
m/s 2 .<br />
Sol. m p = 20 g m Q = 40 g<br />
The horizontal velocities of both the particles are<br />
same and since both are projected simultaneously,<br />
these particle will meet exactly in the middle of AB<br />
(horizontally).<br />
To find the vertical velocity at the time of collision<br />
let us consider the motion of P in vertical and<br />
horizontal directions.<br />
49<br />
49<br />
m / s<br />
m / s<br />
2<br />
2<br />
P 49m/s Q<br />
49m/s<br />
135º<br />
45º<br />
A 49 / 2m / s 49 / 2m / s B<br />
Horizontal direction<br />
S x = 122.5<br />
T x = ?<br />
v x = 49 2<br />
∴ velocity =<br />
⇒<br />
49 122.5 =<br />
2 t x<br />
245m<br />
(122.5) 2<br />
∴ t x =<br />
49<br />
Vertical Direction v y = ?<br />
49<br />
u y =<br />
2<br />
(122.5) 2<br />
t y =<br />
49<br />
a y = – 9.8 m/s 2<br />
displacement<br />
time<br />
49 (122.5)<br />
∴ v y = u y + at y = – 9.8 ×<br />
2<br />
49<br />
m p V p m Q V Q – +<br />
Before collision<br />
m p V p m Q V Q<br />
′<br />
After collision<br />
Also, v 2 – u 2 = 2as<br />
⎛ 49 ⎞<br />
∴ –<br />
⎜<br />
3<br />
⎟<br />
⎝ ⎠<br />
2<br />
⇒ s = 61.25<br />
= 2 (– 9.8) × s<br />
2<br />
= 0<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 10 MAY 2010
⇒ The collision takes place at the maximum<br />
height where the velocities of both the particles will<br />
be in the horizontal direction.<br />
Applying conservation of linear momentum in the<br />
horizontal direction with the information that P<br />
retraces its path therefore its momentum will be same<br />
in magnitude but different in direction.<br />
Momentum of system after collision<br />
∴ m p m p – m Q v Q = – m P v P + m Q v′ Q<br />
2mPvP<br />
− mQvQ<br />
∴ v′ Q =<br />
m<br />
2×<br />
0.02×<br />
(49 / 2) − 0.040(49 /<br />
=<br />
0.040<br />
49 ⎡ 0.04 ⎤ 49<br />
= ⎢ −1⎥ = × 0 = 0<br />
2 ⎣0.040<br />
⎦ 2<br />
New Path of Q after Collision<br />
Considering vertical Motion of Q<br />
u y = 0<br />
s y = – 61.25<br />
a y = – 9.8<br />
t y = ?<br />
Q<br />
2)<br />
S = ut + 2<br />
1 at 2 = 2<br />
1 × (– 9.8) × t 2 = (–61.25)<br />
∴ t = 3.53 sec<br />
Considering Horizontal motion of Q :<br />
'<br />
Since the V Q = 0, therefore the particle Q falls down<br />
vertically so it falls down on the mid point of AB.<br />
3. Three particles, each of mass m, are situated at the<br />
vertices of an equilateral triangle of side length a.<br />
The only forces acting on the particles are their<br />
mutual gravitational forces. It is desired that each<br />
particle moves in a circle while maintaining the<br />
original mutual separation a. Find the initial velocity<br />
that should be given to each particle and also the time<br />
period of the circular motion. [<strong>IIT</strong>-1988]<br />
Sol. The radius of the circle<br />
2<br />
r =<br />
3<br />
a<br />
=<br />
3<br />
a<br />
2 −<br />
2<br />
a<br />
4<br />
m<br />
Let v be the velocity given. The centripetal force is<br />
provided by the resultant gravitational attraction of<br />
the two masses<br />
2 2 2<br />
F R = F + F + 2F cos60º<br />
m× m<br />
= 3 F = 3 G<br />
2<br />
a<br />
2<br />
m mv 2<br />
∴ 3 G<br />
a 2 = r<br />
⇒ v 2 3G ma<br />
=<br />
a 3<br />
2 ×<br />
Gm<br />
⇒ v =<br />
a<br />
Time period of circular motion<br />
2π r 2πa 3<br />
T = = = 2π<br />
v Gm<br />
a<br />
a 3<br />
3Gm<br />
4. Two fixed charges –2Q and Q are located at the points<br />
with coordinates (–3a, 0) and (+3a, 0) respectively in<br />
the x-y plane.<br />
[<strong>IIT</strong>-1991]<br />
(a) Show that all points in the x-y plane where the<br />
electric potential due to the two charges is zero, lie on<br />
a circle. Find its radius and the location of its centre.<br />
(b) Give the expression V(x) at a general point on the x-axis<br />
and sketch the function V(x) on the whole x-axis.<br />
(c) If a particle of charge +q starts form rest at the centre<br />
of the circle, show by a short quantative argument<br />
that the particle eventually crosses the circle. Find its<br />
speed when it does so.<br />
Sol. (a) Let P be a point in the X-Y plane with coordinates<br />
(x, y) at which the potential due to charges<br />
–2Q and +Q placed at A and B respectively be zero.<br />
Y<br />
P(x,y)<br />
∴<br />
A<br />
B<br />
X′ (–3a,0) O x +Q C<br />
(–3a,0)<br />
(5a,0)<br />
(3a+x) (3a-x)<br />
Y′<br />
K(2Q)<br />
=<br />
K( + Q)<br />
2 2<br />
(3a + x) + y<br />
2 2<br />
(3a − x) + y<br />
y<br />
X<br />
m<br />
F<br />
v<br />
F R<br />
F<br />
a/ 3<br />
a<br />
m<br />
⇒ 2<br />
2<br />
2<br />
( 3a − x) + y =<br />
2<br />
( 3a + x) + y<br />
⇒ 4[(3a – x) 2 + y 2 ] = [(3a + x) 2 + y 2 ]<br />
⇒ 4[6a 2 + x 2 – 12ax + y 2 ] = [6a 2 + x 2 + 12ax + y 2 ]<br />
⇒ 3x 2 + 3y 2 – 30ax + 27a 2 = 0<br />
⇒ x 2 + y 2 – 10ax + 9a 2 = 0<br />
⇒ (x – 5a) 2 + (y – 0) 2 = (4a) 2<br />
2<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 11 MAY 2010
This is the equation of a circle with centre at (5a, 0)<br />
and radius 4a. Thus c (5a, 0) is the centre of the<br />
circle.<br />
(b) For x > 3a<br />
To find V(x) at any point on X-axis let us consider a<br />
point (arbitrary) M at a distance x from the origin.<br />
–2Q<br />
+Q<br />
(–3a,0) O (3a,x) M<br />
x<br />
The potential at M will be<br />
(c)<br />
K( −2Q)<br />
K( + Q)<br />
V(x) = +<br />
x + 3a (x − 3a)<br />
where k =<br />
1<br />
4πε<br />
⎡ 1 2 ⎤<br />
∴ V(x) = KQ ⎢ − ⎥ For |x| > 3a<br />
⎣ x − 3a x + 3a<br />
⎦<br />
Similarly, for<br />
⎡ 1 2 ⎤<br />
0 < |x| < 3a V(x) = KQ ⎢ − ⎥<br />
⎣3a<br />
− x 3a + x ⎦<br />
Since circle of zero potential cuts the x-axis at (a, 0)<br />
and (9a, 0) hence V(x) = 0 at x = a at x = 9a<br />
• From the above expressions<br />
V(x) → ∞ at x → 3a and V(x) → – ∞ and<br />
x → – 3a.<br />
• V(x) → 0 as x → + ∞<br />
1<br />
• V(x) varies at in general. x<br />
V<br />
–3a<br />
X<br />
a 3a<br />
Applying Energy Conservation<br />
(K.E. + P.E.) centre = (KE. + P.E.) circumference<br />
⎡Qq<br />
2Qq ⎤ 1<br />
0 + K ⎢ − ⎥ = mv 2 ⎡Qq<br />
2Qq ⎤<br />
+ K<br />
⎣ 2a 8a ⎦ 2<br />
⎢ − ⎥<br />
⎣ 6a 12a ⎦<br />
⇒ 2<br />
1 mv -2 =<br />
KQq<br />
4a<br />
⇒ v =<br />
KQq =<br />
2ma<br />
1<br />
4πε<br />
0<br />
0<br />
⎛ Qq ⎞<br />
⎜ ⎟⎠<br />
⎝ 2ma<br />
5. A thin uniform wire AB of length 1m, an unknown<br />
resistance X and a resistance of 12Ω are connected<br />
by thick conducting strips, as shown in the figure. A<br />
battery and a galvanometer (with a sliding jockey<br />
connected to it) are also available. Connections are to<br />
be made to measure the unknown resistance X using<br />
the principle of Wheatstone bridge. Answer the<br />
following questions.<br />
[<strong>IIT</strong>-2002]<br />
X 12Ω<br />
A B C D<br />
(a) Are there positive and negative terminals on the<br />
galvanometer ?<br />
(b) Copy the figure in your answer book and show the<br />
battery and the galvanometer (with jockey) connected<br />
at appropriate points.<br />
(c) After appropriate connections are made, it is found<br />
that no deflection takes place in the galvanometer<br />
when the sliding jockey touches the wire at a distance<br />
of 60 cm from A. Obtain the value of the resistance<br />
of X.<br />
Sol. (a) No. There are no positive and negative terminals<br />
on the galvanometer.<br />
R AJ 0.6ρ<br />
12Ω<br />
(b) & (c) Q Bridge is balanced = =<br />
R 0.4ρ<br />
X<br />
⇒ X = 8 Ω<br />
where ρ is the resistance per unit length.<br />
A<br />
J<br />
G<br />
X<br />
C<br />
JB<br />
12Ω<br />
CHEMISTRY<br />
6. The oxides of sodium and potassium contained in a<br />
0.5 g sample of feldspar were converted to the<br />
respective chlorides. The weight of the chlorides thus<br />
obtained was 0.1180 g. Subsequent treatment of the<br />
chlorides with silver nitrate gave 0.2451 g of silver<br />
chloride. What is the percentage of Na 2 O and K 2 O in<br />
the mixture ?<br />
[<strong>IIT</strong>-1979]<br />
Sol. Mass of sample of feldspar containing Na 2 O and<br />
K 2 O = 0.5 g.<br />
According to the question,<br />
Na 2 O + 2HCl → 2NaCl + H 2 O ..(1)<br />
2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g<br />
K 2 O + 2HCl → 2KCl + H 2 O ...(2)<br />
2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g<br />
Mass of chlorides = 0.1180 g<br />
Let, Mass of NaCl = x g<br />
∴ Mass of KCl = (0.1180 – x)g<br />
Again, on reaction with silver nitrate,<br />
NaCl + AgNO 3 → AgCl + NaNO 3 ...(3)<br />
23 + 35.5 = 58.5g 108 + 35.5 = 143.5g<br />
KCl + AgNO 3 → AgCl + KNO 3 ...(4)<br />
39 + 35.5 = 74.5g 108 + 35.5 = 143.5g<br />
Total mass of AgCl obtained = 0.2451 g<br />
Step 1. From eq. (3)<br />
D<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 12 MAY 2010
58.5 g of NaCl yields = 143.5 g AgCl<br />
143.5<br />
∴ x g of NaCl yields = x g AgCl<br />
58.5<br />
And from eq. (4),<br />
74.5 g of KCl yields = 143.5 g of AgCl<br />
∴ (0.1180 – x)g of KCl yields<br />
143.5<br />
= (0.1180 – x)g AgCl<br />
74.5<br />
Total mass of AgCl<br />
143.5 143.5<br />
x + (0.1180 – x) = 0.2451<br />
58.5 74.5<br />
which gives, x = 0.0342<br />
Hence, Mass of NaCl = x = 0.0342 g<br />
And Mass of KCl = 0.1180 – 0.0342 = 0.0838g<br />
Step 2. From eq.(1),<br />
117 g of NaCl is obtained from = 62 g Na 2 O<br />
∴ 0.0342 g NaCl is obtained from<br />
62<br />
= × 0.032 = 0.018 g Na2 O<br />
117<br />
From eq. (2),<br />
149 g of KCl is obtained from = 94 g K 2 O<br />
∴ 0.0838 g of KCl is obtained from<br />
94<br />
= × 0.0838 = 0.053 g K2 O<br />
149<br />
0.018<br />
Step 3. % of Na 2 O in feldspar = × 100 = 3.6%<br />
0.5<br />
0.053<br />
% of K 2 O in feldspar = × 100 = 10.6 %<br />
0.5<br />
7. A metallic element crystallizes into a lattice<br />
containing a sequence of layers of ABABAB ..... Any<br />
packing of spheres leaves out voids in the lattice.<br />
What percentage by volume of this lattice is empty<br />
space ?<br />
[<strong>IIT</strong>-1996]<br />
Sol. A unit cell of hcp structure is a hexagonal cell, which<br />
is shown in fig. Three such cells form one hcp unit.<br />
For hexagonal cell, a = b ≠ c; α = β = 90º and<br />
γ = 120º. It has 8 atoms at the corners and one inside,<br />
hence<br />
Number of atoms per unit cell = 8<br />
8 + 1 = 2<br />
O<br />
a<br />
60º<br />
N b<br />
3<br />
Area of the base = b × ON = b × a sin 60º = a<br />
2<br />
2<br />
( Q b = a)<br />
Volume of the hexagonal cell<br />
3<br />
= Area of the base × height = a 2 . c<br />
2<br />
But c =<br />
2 2<br />
a<br />
3<br />
c<br />
β α<br />
b<br />
a γ<br />
∴ Volume of the hexagonal cell<br />
3<br />
= a 2 2 2<br />
. a = a 3 2<br />
2 3<br />
and radius of the atom,<br />
r = a/2<br />
Hence, fraction of total volume of atomic packing<br />
Volume of 2 atoms<br />
factor =<br />
Volume of the hexagonal cell<br />
4 4 ⎛ a ⎞<br />
3<br />
2×<br />
πr<br />
2 × π⎜<br />
⎟<br />
=<br />
3 3 2<br />
=<br />
⎝ ⎠ π<br />
=<br />
3<br />
3<br />
a 2 a 2 3 2<br />
= 0.74 = 74%<br />
∴ The percentage of void space = 100 – 74 = 26%<br />
8. A basic nitrogen compound gave a foul smelling gas<br />
when treated with CHCl 3 and alc. KOH. A 0.295 g<br />
sample of the substance, dissolved in aq. HCl and<br />
treated with NaNO 2 solution at 0ºC liberated a<br />
colourless, odourless gas whose volume corresponds<br />
to 112 ml at STP. After the evolution of gas was<br />
completed, the aq. solution was distilled to give an<br />
organic liquid which did not contain nitrogen and<br />
which on warming with alkali and I 2 gave a yellow<br />
precipitate. Identify the original sustance. Assume<br />
that it contains one N atom per molecule. [<strong>IIT</strong>-1993]<br />
Sol. As the compound on heating with CHCl 3 and alc.<br />
KOH gives foul smelling gas, it should be any<br />
primary amine.<br />
RNH 2 + CHCl 3 + 3KOH<br />
3<br />
⎯⎯→<br />
∆ RN C<br />
(Alkyl isocyanide<br />
(foul smelling gas)<br />
+ 3KCl + 3H 2 O<br />
Since the compound on treating with NaNO 2 and HCl<br />
at 0ºC produces a colourless gas, the compound must<br />
be a p-aliphatic amine, because if it was aromatic<br />
diazonium salt might have been produced<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 13 MAY 2010
NaNO 2 + HCl → NaCl + HNO 2<br />
RNH 2 + HNO 2 → ROH + N 2 + H 2 O<br />
Thus, the gas liberated is N 2 .<br />
Amount of gas produced =<br />
112 ml<br />
22,400 ml<br />
=<br />
1<br />
200<br />
From the above equation, it is obvious that amount of<br />
1<br />
compound RNH 2 = mol. 200<br />
If M is the molar mass of RNH 2 , then<br />
0.295g<br />
Molar mass (M)<br />
1<br />
= 200<br />
mol<br />
ml<br />
∴ M = 0.295 × 200 g mol –1 = 59 g mol –1<br />
Thus, the molar mass of alkyl group (R—) will be,<br />
59 – 16 = 43 g mol –1 .<br />
Hence, R = C 3 H 7 —, i.e., CH 3 CH 2 CH 2 –<br />
or CH 3<br />
CH–<br />
CH 3<br />
The original compound may be either<br />
CH 3 CH 2 CH 2 NH 2 or CH 3 – CH – CH 3<br />
NH 2<br />
From equation, it is clear that the liquid obtained<br />
after distillation is ROH. Since it gives yellow ppt.<br />
with NaOH and I 2 , it must have CH 3 — C — group.<br />
OH<br />
Hence, it is concluded that ROH is CH 3 – CH – CH 3<br />
.<br />
OH<br />
Thus, the original compound is CH 3 – CH – CH 3<br />
.<br />
NH 2<br />
The different equations are :<br />
CH 3<br />
CH 3<br />
CH– NH 2 + CHCl 3 + 3KOH<br />
∆<br />
Isopropyl amine<br />
CH 3<br />
CH 3<br />
CH – N C + 3KCl +<br />
Isopropyl isocyanide<br />
CH 3<br />
CH 3<br />
CH – NH 2 + HNO 2 ⎯→<br />
CH 3<br />
CH – OH + N 2 + H 2 O<br />
CH 3<br />
Isopropyl alcohol<br />
CH 3 – CH – CH 3 + I 2 + 2NaOH ⎯→<br />
OH<br />
CH 3 – C – CH 3<br />
O<br />
Acetone<br />
CH 3 – C – CH 3 + 3I 2 + 3NaOH ⎯→<br />
O<br />
CI 3 – C – CH 3<br />
O<br />
+ 2NaI + 2H 2 O<br />
+ 3NaI + 3H 2 O<br />
∆<br />
CI 3 – C – CH 3 + NaOH ⎯→ CH 3 COONa + CHI 3<br />
Yellow ppt.<br />
O<br />
9. An organic compound (A) C 8 H 6 , on treatment with<br />
dil. H 2 SO 4 containing HgSO 4 gives a compound (B),<br />
which can also be obtained from a reaction of<br />
benzene with an acid chloride in the presence of<br />
anhydrous AlCl 3 . The compound (B) when treated<br />
with iodine in aq. KOH, yields (C) and a yellow<br />
compound (D). Identify (A), (B), (C) and (D) with<br />
justification. Show, how (B) is formed from (A).<br />
[<strong>IIT</strong>-1994]<br />
Sol. The given reactions may be formulated as follows :<br />
C 8 H 6<br />
(A)<br />
Dil H 2SO 4<br />
HgSO 4<br />
AlCl<br />
(B)<br />
3<br />
∆<br />
∆ I 2 + KOH<br />
C 6 H 6 + Acid chloride<br />
(C) + (D)<br />
The reaction of compound (B) with I 2 in KOH is<br />
iodoform reaction. The compound (B) must have a<br />
–COCH 3 group so as to exhibit iodoform reaction.<br />
Since (B) is obtained from benzene by Friedal-Crafts<br />
reaction, it is an aromatic ketone (C 6 H 5 COCH 3 ). The<br />
compound (C) must be potassium salt of an acid.<br />
The compound (A) may be represented as C 6 H 5 C 2 H.<br />
Since it gives C 6 H 5 COCH 3 on treating with dil.<br />
H 2 SO 4 and HgSO 4 , it must contain a triple bond<br />
(–C ≡ CH) in the side chain. Here, the given reactions<br />
may be formulated as follows :<br />
OH<br />
C≡CH<br />
dil H 2SO 4<br />
HgSO 4; H 2O<br />
(A)<br />
C = CH 2<br />
CH 3COCl<br />
AlCl 3; – HCl<br />
Benzene<br />
COCH 3<br />
Acetophenone<br />
(B)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 14 MAY 2010
(B)<br />
Hence. (A)<br />
(C)<br />
O<br />
– C – CH 3<br />
+ 3I 2 + 4KOH<br />
C≡CH<br />
Phenyl acetylene<br />
COOK<br />
Potassium benzoate<br />
∆<br />
–3KI;–3H 2O<br />
COOK<br />
+ CHI3<br />
(D)<br />
(C)<br />
Potassium benzoate<br />
COCH 3<br />
(B)<br />
(D)<br />
Acetophenone<br />
CHI 3<br />
Idoform<br />
10. When 20.02 g of a white solid (X) is heated, 4.4 g of<br />
an acid gas (A) and 1.8g of a neutral gas (B) are<br />
evolved leaving behind a solid residue (Y) of weight<br />
13.8 g. Gas (A) turns lime water milky and (B)<br />
condenses into a liquid which changes anhydrous<br />
CuSO 4 blue. The aqueous solution of (Y) is alkaline<br />
to litmus and gives 19.7 g of a white precipitate (Z)<br />
with BaCl 2 solution. (Z) gives CO 2 with an acid.<br />
Identify (A), (B), (X), (Y) and (Z). [<strong>IIT</strong>-1989]<br />
Sol. (i) Since acidic gas (A) turns lime water milky hence<br />
it is CO 2 or SO 2 , both of which form white insoluble<br />
compound with Ca(OH) 2<br />
(ii) Neutral gas (B) condenses into a liquid which<br />
turns anhydrous CuSO 4 (white) into blue<br />
(CuSO 4 .5H 2 O), hence (B) is H 2 O.<br />
(iii) (Y) gives alkaline solution and its solution forms<br />
white precipitate (Z) with BaCl 2 solution. (Z) on<br />
heating gives the acid gas CO 2 , hence (Z) is BaCO 3<br />
and therefore (Y) is a metal carbonate.<br />
(iv) Since (Y) and (A) are produced from (X), thus<br />
(X) is a metal bicarbonate.<br />
( X) →<br />
20.02 g<br />
( A) +<br />
4.4 g<br />
( B) +<br />
1.80g<br />
( Y)<br />
13.8g<br />
From the above values we may write a general<br />
equation for a bicarbonate.<br />
⎯ ∆ 2<br />
(A)<br />
2MHCO 3 ⎯→<br />
(X)<br />
CO + H 2 O +<br />
(B)<br />
M 2CO 3<br />
(Y)<br />
Q 4.4g CO 2 is obtained from 20.02 g of MHCO 3<br />
∴ 4g CO 2 is obtained from 200.2 g of MHCO 3<br />
200.2<br />
Molecular weight of MHCO 3 = = 100.1<br />
2<br />
Atomic weight of M = 39<br />
Thus, the metal M is potassium and then (X) is<br />
KHCO 3 . The equations are :<br />
⎯ ∆ (Y)<br />
2 KHCO 3 ⎯→<br />
(X)<br />
K 2CO 3 +<br />
K 2CO 3 + BaCl 2 ⎯→ 2KCl +<br />
(Y)<br />
BaCO<br />
(Z)<br />
3<br />
⎯⎯→<br />
∆ BaO +<br />
CO 2 +<br />
(A)<br />
BaCO<br />
(Z)<br />
CO2<br />
↑<br />
(A)<br />
3<br />
H 2O<br />
(B)<br />
Hence, (A) is CO 2 (B) is H 2 O (X) is KHCO 3 (Y) is<br />
K 2 CO 3 and (Z) is BaCO 3 .<br />
MATHEMATICS<br />
11. From a point A common tangents are drawn to the<br />
circle x 2 + y 2 = a 2 /2 and parabola y 2 = 4ax. Find the<br />
area of the quadrilateral formed by the common<br />
tangents, the chord of contact of the circle and the<br />
chord of contact of the parabola. [<strong>IIT</strong>-1996]<br />
Sol. Equation of any tangent to the parabola, y 2 = 4ax is<br />
y = mx + a/m.<br />
This line will touch the circle x 2 + y 2 = a 2 /2<br />
A(–a, 0)<br />
If<br />
⇒<br />
⎛ a ⎞<br />
⎜ ⎟⎠<br />
⎝ m<br />
2<br />
=<br />
πx = – a/2<br />
y<br />
B<br />
C<br />
O<br />
a 2 (m 2 + 1)<br />
2<br />
E<br />
D<br />
x = a<br />
1 1<br />
2 = (m 2 + 1) ⇒ 2 = m 4 + m 2<br />
m 2<br />
⇒ m 4 + m 2 – 2 = 0<br />
⇒ (m 2 – 1)(m 2 + 2) = 0<br />
⇒ m 2 – 1 = 0, m 2 = – 2 (which is not possible).<br />
⇒ m = ± 1<br />
Therefore, two common tangents are<br />
y = x + a and y = –x – a<br />
These two intersect at A(–a, 0)<br />
The chord of contact of A(–a, 0) for the circle<br />
x 2 + y 2 = a 2 /2 is<br />
(–a)x + 0.y = a 2 /2 or x = – a/2<br />
and chord of contact of A(–a, 0) for the parabola<br />
y 2 = 4ax is<br />
0.y = 2a(x – a) or x = a<br />
Again length of BC = 2BK<br />
L<br />
x<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 15 MAY 2010
2<br />
2<br />
= 2 OB − OK<br />
= 2<br />
2 2<br />
a a a 2<br />
− = 2<br />
2 4 4<br />
= a<br />
and we know that DE is the latus rectum of the<br />
parabola so its length is 4a.<br />
Thus area of the trapezium<br />
BCDE = 2<br />
1 (BC + DE) (KL)<br />
1 ⎛ 3a ⎞<br />
= (a + 4a) ⎜ ⎟⎠ =<br />
2 ⎝ 2<br />
15a<br />
2<br />
4<br />
12. Let V be the volume of the parallelopiped formed by<br />
the vectors<br />
→<br />
a = a1 î + a 2 ĵ + a 3 kˆ ;<br />
→<br />
b = b1 î + b 2 ĵ + b 3 kˆ<br />
→<br />
c = c1 î + c 2 ĵ + c 3 kˆ<br />
If a r , b r , c r , where r = 1, 2, 3 are non-negative real<br />
3<br />
numbers and ∑ ( a r + br<br />
+ cr<br />
) = 3L. Show that<br />
r=<br />
1<br />
V ≤ L 3 .<br />
[<strong>IIT</strong>-2002]<br />
Sol. V = |<br />
→ →<br />
a .(b × →<br />
c ) | ≤ 2 2<br />
a + a +<br />
2<br />
Now, L =<br />
1 2 a 3<br />
2 2 2 2 2 2<br />
1 + b2<br />
b3<br />
c 1 c2<br />
+ c3<br />
b + + ...(1)<br />
( a1 + a 2 + a 3)<br />
+ (b1<br />
+ b2<br />
+ b3)<br />
+ (c1<br />
+ c2<br />
+ c3)<br />
3<br />
[(a 1 + a 2 + a 3 ) (b 1 + b 2 + b 3 ) (c 1 + c 2 + c 3 )] 1/3<br />
∴ L 3 ≥ [(a 1 + a 2 + a 3 )(b 1 + b 2 + b 3 )(c 1 + c 2 + c 3 )] ..(2)<br />
Now, (a 1 + a 2 + a 3 ) 2<br />
=<br />
2<br />
a 1 + a 2 2 + a 3<br />
2 + 2a 1 a 2 + 2a 1 a 3 + 2a 2 a 3 ≥<br />
⇒ (a 1 + a 2 + a 3 ) ≥<br />
Similarly,<br />
and<br />
2 2 2<br />
1 + a 2 a 3<br />
a +<br />
(b 1 + b 2 + b 3 ) ≥<br />
(c 1 + c 2 + c 3 ) ≥<br />
2 2 2<br />
1 + b2<br />
b3<br />
b +<br />
2 2 2<br />
1 + c2<br />
c3<br />
c +<br />
2<br />
a 1 + a 2 2 + a 3<br />
2<br />
∴ from (1) and (2)<br />
L 3 2<br />
≥ [( a 1 + a 2 2 + a 3<br />
2 2 2 2 2 2 2<br />
)( b 1 + b2<br />
+ b3<br />
)( c 1 + c2<br />
+ c3<br />
)] 1/3 ≥ V<br />
13. T is a prallelopiped in which A, B, C and D are<br />
vertices of one face and the just above it has<br />
corresponding vertices A´, B´, C´, D´, T is now<br />
compressed to S with face ABCD remaining same<br />
and A´, B´, C´, D´ shifted to A´´, B´´, C´´, D´´ in S.<br />
The volume of parallelopiped S is reduced to 90% of<br />
T. Prove that locus of A´´ is a plane. [<strong>IIT</strong>-2004]<br />
Sol. Let the equation of the plane ABCD be<br />
ax + by + cz + d = 0, the point A´´ be (α, β, γ) and<br />
the height of the parallelopiped ABCD be h.<br />
⇒<br />
| aα + bβ + cγ + d |<br />
= 90%. h<br />
2 2 2<br />
a + b + c<br />
⇒ aα + bβ + cγ + d = ± 0.9h<br />
2<br />
2<br />
a + b + c<br />
⇒ locus is, ax + by + cz + d = ±0.9h<br />
2<br />
2<br />
2<br />
a + b + c<br />
⇒ locus of A´ is a plane parallel to the plane ABCD<br />
14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6,<br />
is thrown n times and the list on n numbers showing<br />
up is noted. What is the probability that among the<br />
numbers 1, 2, 3, 4, 5, 6 only three numbers appear in<br />
this list ?<br />
[<strong>IIT</strong>-2001]<br />
Sol. Let us define at onto function F from A : [r 1 , r 2 ... r n ]<br />
to B : [1, 2, 3] where r 1 r 2 .... r n are the readings of n<br />
throws and 1, 2, 3 are the numbers that appear in the<br />
n throws.<br />
Number of such functions,<br />
M = N – [n(1) – n(2) + n(3)]<br />
where N = total number of functions and<br />
n(t) = number of function having exactly t elements<br />
in the range.<br />
Now, N = 3 n , n(1) = 3.2 n , n(2) = 3, n(3) = 0<br />
⇒ M = 3 n – 3.2 n + 3<br />
Hence the total number of favourable cases<br />
= (3 n – 3.2 n + 3). 6 C 3<br />
⇒ required probability =<br />
n<br />
n<br />
( 3 − 3.2 + 3) ×<br />
15. A straight line L through the origin meets the line<br />
x + y = 1 and x + y = 3 at P and Q respectively.<br />
Through P and Q two straight lines L 1 and L 2 are<br />
drawn, parallel to 2x – y = 5 and 3x + y = 5<br />
respectively. Lines L 1 and L 2 intersect at R, shown<br />
that the locus of R as L varies, is a straight line.<br />
[<strong>IIT</strong>-2002]<br />
Sol. Let the equation of straight line L be y = mx<br />
⎛<br />
P ≡ ⎜<br />
⎝<br />
1<br />
m<br />
m ⎞ ⎛ 3 3m ⎞<br />
, ⎟ ; Q ≡ ⎜ , ⎟<br />
+ 1 m + 1 ⎠ ⎝ m + 1 m + 1 ⎠<br />
m − 2<br />
Now equation of L 1 : y – 2x = ...(1)<br />
m + 1<br />
3m + 9<br />
equation of L 2 : y + 3x =<br />
...(2)<br />
m + 1<br />
By eliminating 'm' from equation (1) and (2), we get<br />
locus of R as x – 3y + 5 = 0, which represents a<br />
straight line.<br />
6<br />
n<br />
6<br />
C<br />
3<br />
2<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 16 MAY 2010
Physics Challenging Problems<br />
Set # 1<br />
This section is designed to give <strong>IIT</strong> <strong>JEE</strong> aspirants a thorough grinding & exposure to variety<br />
of possible twists and turns of problems in physics that would be very helpful in facing <strong>IIT</strong><br />
<strong>JEE</strong>. Each and every problem is well thought of in order to strengthen the concepts and we<br />
hope that this section would prove a rich resource for practicing challenging problems and<br />
enhancing the preparation level of <strong>IIT</strong> <strong>JEE</strong> aspirants.<br />
By : Dev Sharma<br />
Solutions will be published in next issue<br />
Director Academics, Jodhpur Branch<br />
1. Two capacitors C 1 and C 2 , can be charged to a<br />
potential V/2 each by having<br />
C 1 C 2<br />
O<br />
V R R<br />
S 1 S 2<br />
(A) S 1 closed and S 2 open<br />
(B) S 1 open and S 2 closed<br />
(C) S 1 and S 2 both closed<br />
(D) cannot be charged at V/2<br />
2. Energy liberated in the de-excitation of hydrogen<br />
atom from 3 rd level to 1 st level falls on a photocathode.<br />
Later when the same photo-cathode is<br />
exposed to a spectrum of some unknown<br />
hydrogen like gas, excited to 2 nd energy level, it is<br />
found that the de-Broglie wavelength of the<br />
fastest photoelectrons, now ejected has decreased<br />
by a factor of 3. For this new gas, difference of<br />
energies of 2 nd Lyman line and 1 st Balmer line if<br />
found to be 3 times the ionization potential of the<br />
hydrogen atom. Select the correct statement(s)<br />
(A) The gas is lithium<br />
(B) The gas is helium<br />
(C) The work function of photo-cathode is 8.5eV<br />
(D) The work function of photo-cathode is 5.5eV<br />
3. In the figure shown there exists a uniform time<br />
varying magnetic field B = [(4T/s) t + 0.3T] in a<br />
cylindrical region of radius 4m. An equilateral<br />
triangular conducting loop is placed in the<br />
magnetic field with its centroide on the axis of the<br />
field and its plane perpendicular to the field.<br />
+<br />
+<br />
+<br />
+<br />
B<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
A<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+ +<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
C<br />
(A) e.m.f. induced in any one rod is 16V<br />
(B) e.m.f. induced in the complete ∆ ABC is<br />
48<br />
3V<br />
(C) e.m.f. induced in the complete<br />
(D) e.m.f. induced in any one rod is 16<br />
∆ ABC is 48V<br />
3V<br />
4. 6 parallel plates are arranged as shown. Each<br />
plate has an area A and Distance between them is<br />
as shown. Plate 1-4 and plates 3-6 are connected<br />
equivalent capacitance across 2 and 5 can be<br />
nA ∈<br />
writted as 0 . Find min value of n. (n, d are<br />
d<br />
natural numbers)<br />
1<br />
2 d<br />
3 d<br />
2d<br />
4<br />
d 5<br />
d 6<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 17 MAY 2010
5. Match the following<br />
Column – I<br />
Column – II<br />
(A) A light conducting (P) Magnetic field B<br />
circular flexible is doubled.<br />
loop of wire of<br />
radius r carrying<br />
current I is placed<br />
in uniform magnetic<br />
field B, the tension<br />
in the loop is doubled<br />
if<br />
(B) Magnetic field at a (Q) Inductance is<br />
point due to a long increased by four<br />
straight current times.<br />
carrying wire at a<br />
point near the wire<br />
is doubled if<br />
(C) The energy stored (R) Current I is<br />
in the inductor will doubled<br />
become four times<br />
(D) The force acting on a (S) Radius r is<br />
moving charge, doubled<br />
moving in a constant<br />
magnetic field will be<br />
doubled if<br />
(T) Velocity v is<br />
Doubled<br />
Passage # (Q. No. 6 to Q. No. 8 )<br />
A solid, insulating ball of radius ‘a’ is surrounded<br />
by a conducting spherical shell of inner radius ‘b’<br />
and outer radius ‘c’ as shown in the figure. The<br />
inner ball has a charge Q which is uniformly<br />
distribute throughout is volume. The conducting<br />
spherical shell has a charge –Q.<br />
Answer the following questions.<br />
–Q<br />
b c<br />
Q<br />
a<br />
6. Assuming the potential at infinity to be zero, the<br />
potential at a point located at a distance a/2 from<br />
the centre of the sphere will be<br />
Q ⎡2<br />
1 ⎤<br />
Q ⎡11<br />
1 ⎤<br />
(A) ⎢ −<br />
4πε<br />
⎥ (B)<br />
⎣a<br />
b<br />
⎢ − ⎥ ⎦ 4πε<br />
⎣8a<br />
b ⎦<br />
(C)<br />
Q<br />
4πε<br />
0<br />
0<br />
⎡1<br />
1 ⎤<br />
⎢ − ⎥<br />
⎣a<br />
b ⎦<br />
0<br />
(D) None of these<br />
7. Work done by external agent in taking a charge q<br />
slowly from inner surface of the shell to surface<br />
of the sphericalball will be<br />
⎡1<br />
1⎤<br />
⎡1<br />
1⎤<br />
(A) kQq ⎢ − ⎥ (B) kQq<br />
⎣a<br />
c<br />
⎢ − ⎥ ⎦ ⎣b<br />
a ⎦<br />
⎡1<br />
1 ⎤<br />
⎡1<br />
1⎤<br />
(C) kQq ⎢ − ⎥ (D) kQq<br />
⎣a<br />
b<br />
⎢ − ⎥ ⎦ ⎣c<br />
a ⎦<br />
8. Now the outer shell is grounded, i.e., the outer<br />
surface is fixed to be zero. Now the charge on the<br />
inner ball will be<br />
(A) zero<br />
(B) Q<br />
(C)<br />
Q ⎛ 1<br />
⎜ +<br />
C ⎝ a<br />
1<br />
c<br />
1 ⎞<br />
− ⎟<br />
b ⎠<br />
(D)<br />
Q ⎛ 1<br />
⎜ +<br />
b ⎝ a<br />
1<br />
c<br />
Cartoon Law of Physics<br />
1 ⎞<br />
− ⎟<br />
b ⎠<br />
Any body passing through solid matter will leave a<br />
perforation conforming to its perimeter.<br />
Also called the silhouette of passage, this<br />
phenomenon is the specialty of victims of directedpressure<br />
explosions and of reckless cowards who<br />
are so eager to escape that they exit directly<br />
through the wall of a house, leaving a cookiecutout-perfect<br />
hole. The threat of skunks or<br />
matrimony often catalyzes this reaction.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 18 MAY 2010
PHYSICS<br />
Students'Forum<br />
Expert’s Solution for Question asked by <strong>IIT</strong>-<strong>JEE</strong> Aspirants<br />
⎡ m ⎤<br />
1. A fighter plane flies at a velocity of 300 ⎢ ⎥ . On<br />
⎣sec<br />
⎦<br />
the fighter plane there is a gun which shoots at a rate<br />
of 40 rounds per second with a muzzle velocity of<br />
⎡ m ⎤<br />
1200 ⎢ ⎥ . The shots are aimed at another fighter<br />
⎣sec<br />
⎦<br />
⎡ m ⎤<br />
plane flying at a velocity of 200 ⎢ ⎥ . Find the rate<br />
⎣sec<br />
⎦<br />
at which the projectiles hit the target plane :<br />
(a) When the two planes move in the same direction, and<br />
the target plane is in front of the shooting plane.<br />
(b) The same as (a), when the target plane is in the rear<br />
of the shooting plane.<br />
(c) When the two planes move towards one another.<br />
(d) When the two planes move away from one another.<br />
Sol. Denote by v s the velocity of the plane from which the<br />
shots are fired, by v t the velocity of the target plane<br />
and by L the distance between them at the certain<br />
moment of time when the shooting plane starts to<br />
shoot. Denote by r the rate of fire of the gun and by v<br />
the muzzle velocity.<br />
(a) The time it takes for the first projectile to reach the<br />
target plane is:<br />
L<br />
t 1 =<br />
…(1)<br />
v + v s − v t<br />
1<br />
After a time of the second projectile is shot, and<br />
r<br />
the distance between the planes at this time is given<br />
by:<br />
vs vt<br />
L' = L –<br />
−<br />
…(2)<br />
r<br />
Thus, the time it takes the second projectile to arrive<br />
at the target plane is:<br />
v s − v t<br />
L −<br />
t 2 =<br />
r<br />
…(3)<br />
v + v s − v t<br />
which is<br />
∆t = t 2 + r<br />
1 – t1<br />
1 v<br />
= –<br />
s − vt<br />
v<br />
=<br />
…(4)<br />
r r(v + vs<br />
− vt<br />
) r(v + vs − vt<br />
)<br />
after the first shot. Naturally, the time increment does<br />
not depend on the initial distance; thus the rate of<br />
hitting is:<br />
1<br />
r′ =<br />
∆ = r ( v + vs − vt<br />
)<br />
t v<br />
1300 ⎡ hits ⎤<br />
= 40 × = 43.33 1200<br />
⎢ ⎥ … (5)<br />
⎣ second ⎦<br />
(b) Using the same reasoning for this case, we obtain:<br />
v − vs + vt<br />
r′ = r<br />
v<br />
1100 ⎡ hits ⎤<br />
= 40 × = 36.66 1200<br />
⎢ ⎥ … (6)<br />
⎣ second ⎦<br />
(c) In this case:<br />
v + vs + vt<br />
r′ = r<br />
v<br />
1700 ⎡ hits ⎤<br />
= 40 × = 56.67 1200<br />
⎢ ⎥ … (7)<br />
⎣ second ⎦<br />
(d) Here,<br />
v v v<br />
r′ = r<br />
− s − t<br />
v<br />
700 ⎡ hits ⎤<br />
= 40 × = 23.33 1200<br />
⎢ ⎥ … (8)<br />
⎣ second ⎦<br />
2. Consider the system described in figure.<br />
m 1<br />
m 2<br />
(a) Use the equations of energy conservation to find<br />
the velocities of masses m 1 and m 2 after they are<br />
released from rest and pass a distance y<br />
(assume m 2 > m 1 ).<br />
(b) Use the expression obtained in the first section to<br />
find the acceleration of the masses.<br />
Sol. (a) We write the change in the potential energy of the<br />
masses :<br />
m 1 : = ∆E p = m 1 gy ...(1)<br />
m 2 : ∆E p = – m 2 gy ...(2)<br />
The change in the kinetic energy is<br />
∆E k = 2<br />
1<br />
m1 v 1 2 + 2<br />
1<br />
m2 v 2<br />
2<br />
...(3)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 19 MAY 2010
Because | v r 1 | = | v r 2 | ≡ v, we obtain<br />
∆E k = 2<br />
1 (m1 + m 2 )v 2 ...(4)<br />
Since there are no external forces except gravity,<br />
which is a conservative force, we know, using energy<br />
conservation that :<br />
∆E p + ∆E k = 0 ...(5)<br />
By substituting<br />
(m 1 – m 2 )gy + 2<br />
1 (m1 + m 2 )v 2 = 0 ...(6)<br />
1/ 2<br />
⎛ 2g(m2<br />
− m1)<br />
⎞<br />
Hence, v(y) =<br />
⎜<br />
m1<br />
m<br />
⎟<br />
⎝ + 2 ⎠<br />
y ...(7)<br />
(b) By definition,<br />
1/ 2<br />
dv ⎛ 2g(m<br />
a = =<br />
2 − m1)<br />
⎞ d y<br />
dt<br />
⎜<br />
m1<br />
m<br />
⎟ .<br />
⎝ + 2 ⎠ dt<br />
⎛ 2g(m2<br />
− m1)<br />
⎞<br />
=<br />
⎜<br />
m1<br />
m<br />
⎟<br />
⎝ + 2 ⎠<br />
1/ 2<br />
2<br />
1<br />
y<br />
dy<br />
. dt<br />
...(8)<br />
dy<br />
and because = v we substitute Eq. (7) into Eq.<br />
dt<br />
(8) and obtain.<br />
m2<br />
− m1<br />
a = g<br />
m + m<br />
2<br />
1<br />
3. A smooth incline of lift angle α is accelerated at a<br />
rate α. A block of mass m is placed on the incline. At<br />
t = 0 the block is released, and begins moving (see<br />
figure.)<br />
N<br />
a<br />
mg<br />
m<br />
(a) Write the equations of motion for the block.<br />
(b) What is the maximal value of 'a' for which the<br />
block will remain attached to the incline ?<br />
(c) How much time is required for the block to slide<br />
a distance L along the incline ?<br />
(d) If we accelerate the incline in the opposite<br />
direction, what is the minimal value of a necessary<br />
for the block to slide up the incline ?<br />
Sol. (a) D'alembert's force exists in the acclerated system,<br />
on the plane of the incline. Therefore,<br />
m a r = mg xˆ ´ sin α – mg ŷ ´ cos α + N ŷ ´ – m a r<br />
a r is expressed by the unit vectors of the accelerated<br />
system as :<br />
a r = – a xˆ ´ cos α – a ŷ ´ sin α<br />
and in component form :<br />
⎧a<br />
⎪<br />
x´ = g sin α + a cos α<br />
⎨ N<br />
a y´ = − g cos α + a sin α (N ≥ 0)<br />
⎪⎩ m<br />
y´<br />
α<br />
x´<br />
(b) The maximal value is obtained when N = 0 and<br />
a y´ = 0. (The block is still upon the inclined surface,<br />
as stipulated.) Therefore,<br />
a = g cot α<br />
(c) The equation of motion with a constant<br />
acceleration 'a' is :<br />
x(t) = x 0 + v 0 t + 2<br />
1 at<br />
2<br />
1<br />
In our case, we have L = a x ´ t 2 . Therefore,<br />
2<br />
2L<br />
t =<br />
gsin α + a cosα<br />
(d) Because 'a' is changed to (–a), the equation of<br />
motion in the xˆ ´ direction becomes :<br />
a x´ = g sin α – a cos α<br />
In order for the mass to slide up the incline, the<br />
condition a x´<br />
> 0 must be met. Thus,<br />
a > g tan α<br />
4. Two long wires are placed on a smooth horizontal<br />
table. Wires have equal but opposite charges.<br />
Magnitude of linear charge density on each wire is λ.<br />
Calculate (for unit length of wires) work required to<br />
increase the separation between the wires from a to 2a.<br />
Sol. Since, wires have opposite charge, therefore, they<br />
attract each other. To increase separation between the<br />
wires, work is to be done against this force of<br />
attraction.<br />
Let at some instant separation between the wires be x<br />
as shown in Fig.<br />
To calculate force of attraction between the wires,<br />
first electric field due to charge on one wire at<br />
position of the second wire is to be calculated.<br />
Therefore, considering a cylindrical surface of radius<br />
x and of unit length, co-axial with positively charged<br />
wire,<br />
+ –<br />
Its area = 2π x × 1<br />
Charge enclosed within it = λ<br />
+<br />
+<br />
+<br />
x<br />
∴ Flux passing through the cylindrical surface =<br />
Electric field,<br />
–<br />
–<br />
–<br />
E = Flux passing per unit area.<br />
( λ / ε0)<br />
λ<br />
= =<br />
2π<br />
x 2πε0x<br />
Magnitude of charge on unit length of second wire = λ<br />
∴ Force of attraction per unit length is F = λ E<br />
λ<br />
ε 0<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 20 MAY 2010
2<br />
λ<br />
or F =<br />
2πε0x<br />
To increase the separation, wires are to be pulled apart<br />
by applying an infinitesimally greater force (F + dF).<br />
∴ Work done to increase separation from x to (x + dx),<br />
dW = (F + dF) dx ≈ F. dx<br />
or Total work done =<br />
∫ = x=<br />
x<br />
2<br />
a<br />
λ<br />
=<br />
2πε<br />
2a<br />
0<br />
x<br />
F dx =<br />
∫ = 2a λ<br />
x= a 2πε<br />
log e 2<br />
2<br />
0<br />
dx<br />
x<br />
5. Two short electric dipoles having dipole moment p 1<br />
and p 2 are placed co-axially and uni-directionally, at<br />
a distance r apart. Calculate nature and magnitude of<br />
force between them.<br />
Sol. Let second dipole having dipole moment p 2 consist of<br />
charges (+ q 2 ) and (– q 2 ) which are separated by an<br />
elemental distance 2dr as shown in Fig.<br />
Then p 2 = q 2 2dr …(1)<br />
→<br />
p 1<br />
r<br />
2.dr<br />
– +<br />
q 2 q 2<br />
Since, dipoles are separated by a distance r, it means<br />
distance between their centres is r. Distance of<br />
charges (– q 2 ) & (+ q 2 ) from centre of first dipole is<br />
(r – dr) & (r + dr), respectively. If electric field<br />
strength due to and at distance r from dipole having<br />
dipole moment p 1 is E, then electric field strength at<br />
position of two charges will be (E – dE) & (E + dE),<br />
respectively.<br />
1 2p1<br />
Where E =<br />
(rightward)<br />
3<br />
4πε<br />
r<br />
0<br />
∴<br />
1 2p1<br />
dE = – 3 . . dr<br />
4<br />
4πε0<br />
r<br />
Force on charge (– q 2 ) is F 1 = (E – dE) q 2 (leftward)<br />
and that on charge (+ q 2 ) is F 2 = (E + dE) q 2<br />
(rightwards)<br />
Hence, net force on second dipole is<br />
F = F 2 – F 1 (rightward)<br />
or<br />
1 2p1<br />
F = dE 2q 2 = – 3 . dr 2q<br />
4 2<br />
4πε0<br />
r<br />
But 2q 2 dr = p 2<br />
∴<br />
1 6p1p2<br />
Net force = –<br />
4<br />
4πε0<br />
r<br />
(–ve) sign indicates that actual direction of force on<br />
second dipole is leftward or force between<br />
two dipoles is of attraction and its magnitude is<br />
1<br />
F =<br />
4πε 6p1p2<br />
4<br />
r<br />
0<br />
1. The typical size of a meteor is about one<br />
cubic centimeter, which is equivalent to the<br />
size of a sugar cube.<br />
2. Each day, Earth accumulate 10 to 100 tons of<br />
material.<br />
3. There are over 100 billion galaxies in the<br />
universe.<br />
4. The largest galaxies contain nearly 400 billion<br />
stars.<br />
5. The risk of a falling meteorite striking a<br />
human occurs once every 9,300 years.<br />
6. A piece of a neutron star the size of a pin<br />
point would way 1 million tons.<br />
7. Europa, Jupiter’s moon, is completely<br />
covered in ice.<br />
8. Light reflecting off the moon takes 1.2822<br />
seconds to reach Earth.<br />
9. There has only been one satellite destroyed<br />
by a meteor, it was the European Space<br />
Agency’s Olympus in 1993.<br />
10. The International Space Station orbits at 248<br />
miles above the Earth.<br />
11. The Earth orbits the Sun at 66,700mph.<br />
12. Venus spins in the opposite direction<br />
compared to the Earth and most other<br />
planets. This means that the Sun rises in the<br />
West and sets in the East.<br />
13. The Moon is moving away from the Earth at<br />
about 34cm per year.<br />
14. The Sun, composed mostly of helium and<br />
hydrogen, has a surface temperature of 6000<br />
degrees Celsius.<br />
15. A manned rocket reaches the moon in less<br />
time than it took a stagecoach to travel the<br />
length of England.<br />
16. The nearest known black hole is 1,600 light<br />
years (10 quadrillion miles/16 quadrillion<br />
kilometers) away.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 21 MAY 2010
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 22 MAY 2010
PHYSICS FUNDAMENTAL FOR <strong>IIT</strong>-<strong>JEE</strong><br />
Electrostatics-I<br />
KEY CONCEPTS & PROBLEM SOLVING STRATEGY<br />
• Coulomb's Law :<br />
1 q1q<br />
2<br />
F 0 =<br />
(in vacuum)<br />
2<br />
4πε0 r<br />
Vectorially → 1 q1q<br />
2<br />
F= rˆ<br />
2<br />
4πε<br />
0 r<br />
1 q1q<br />
2<br />
In any material medium F =<br />
2<br />
4πε0εr<br />
r<br />
where ε r is a constant of the material medium called<br />
its relative permittivity, and ε 0 is a universal constant,<br />
called the permittivity of free space.<br />
ε 0 = 8.85 × 10 –12 1<br />
or<br />
4πε = 9 × 109<br />
0<br />
The unit of ε 0 is C 2 N m –2 or farad per metre.<br />
1 q1q<br />
2<br />
Also F =<br />
2<br />
4πε0εr<br />
r<br />
Where ε is called the absolute permittivity of the<br />
medium.<br />
Obviously, ε r = F 0 /F. Remember ε r = ∞ for<br />
conductors.<br />
Conductors and insulators Each body contains<br />
enormous amounts of equal and opposite charges. A<br />
'charged' body contains an excess of either positive or<br />
negative charge.<br />
In a conductor, some of the negative charges are free<br />
to move around. In an insulator (also called a<br />
dielectric), the charges cannot move. They can only<br />
undergto small localized displacements, causing<br />
polarization.<br />
Induction When a charged body A is brought near<br />
another body B, unlike charges are induced on the<br />
near surface of B (called bound charges) and like<br />
charges appear on the far surface of B (called free<br />
charges) If B is a conductor, the free charges can be<br />
removed by earthing B, e.g., by touching it. If B is an<br />
insulator, separation of like and unlike charges will<br />
still occur due to induction. However, the like<br />
charges cannot then be removed by earthing B.<br />
• Electric Field And Potential<br />
Electric Field An electric field of strength E is said<br />
to exist at a point if a test charge ∆q at that point<br />
experiences a force given by<br />
→<br />
→ →<br />
→<br />
∆ F<br />
∆ F = ∆q<br />
F or E =<br />
∆q<br />
The unit of electric field is Newton per coulomb or<br />
volt per metre. The electric field strength at a<br />
distance r from a point charge q in a medium of<br />
permittivity ε is given by<br />
E =<br />
1<br />
4πε<br />
q<br />
2<br />
r<br />
Vectorially → 1 q<br />
E = 4πε<br />
2 rˆ<br />
r<br />
With reference to any origin<br />
→<br />
E =<br />
q<br />
4πε<br />
→<br />
R−<br />
r<br />
→<br />
→<br />
3<br />
→<br />
R−<br />
r<br />
Where R → is the position vector of the field point and<br />
→<br />
r , the position vector of q.<br />
Due to a number of discrete charges<br />
i=<br />
N<br />
→<br />
→<br />
→<br />
q E =<br />
1 R−<br />
ri<br />
∑<br />
3<br />
4πε<br />
→ →<br />
i=<br />
1<br />
R−<br />
r<br />
i<br />
Electric Potential The electric potential at a point is<br />
the work done by an external agent in bringing a unit<br />
positive charge from infinity up to that point along<br />
any arbitrary path.<br />
∆W∞→<br />
(by an external agent)<br />
V P = P<br />
volt(V) or JC –1<br />
∆q<br />
The potential difference between two points P and Q<br />
is given by<br />
∆W Q → P (by agent)<br />
V P – V Q =<br />
volt (V)<br />
∆q<br />
The potential at a distance r from a point charge q in<br />
a medium of permittivity ε is<br />
ϕ or V =<br />
1<br />
4πε<br />
q 1 =<br />
r 4πε<br />
q<br />
→ →<br />
R− r<br />
with reference to any arbitrary origin.<br />
Due to a number of charges<br />
i=<br />
N<br />
1 q1<br />
ϕ or V = ∑<br />
4πε<br />
→ →<br />
i= 1 R−<br />
r<br />
i<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 23 MAY 2010
In a conductor, all points have the same potential.<br />
If charge q (coulomb) is placed at a point where the<br />
potential is V (volt), the potential energy of the<br />
system is qV (joule). It follows that if charges q 1 , q 2<br />
are separated by distance r, the mutual potential<br />
q<br />
energy of the system is 1 q 2<br />
4πε r<br />
.<br />
• Relation Between Field (E) And Potential (V)<br />
The negative of the rate of change of potential along<br />
a given direction is equal to the component of the<br />
field that direction.<br />
∂V<br />
E r = – along r<br />
∂ r<br />
∂V and E ⊥ = perpendicular to r<br />
r ∂θ<br />
When two points have different potentials, an electric<br />
field will exist between them, directed from the<br />
higher to the lower potential.<br />
• Lines of Force<br />
A line of force in an electric field is such a curve that<br />
the tangent to it at any point gives the direction of the<br />
field at that point. Lines of force cannot intersect<br />
each other because it is physically impossible for an<br />
electric field to have two directions simultaneously.<br />
• Equipotential Surfaces<br />
The locus of points of equal potential is called an<br />
equipotential surface. Equipotential surfaces lie at<br />
right angles to the electric field. Like lines of force,<br />
they can never intersect.<br />
Note: For solving problems involving electrostatic<br />
units, remember the following conversion factors:<br />
3 × 10 9 esu of charge = 1 C<br />
1 esu of potential = 300 V<br />
• Electric Flux<br />
The electric flux over a surface is the product of its<br />
surface area and the normal component of the electric<br />
field strength on that surface. Thus,<br />
dϕ = (E cos θ) ds = E n ds = → E . → ds<br />
ds<br />
O<br />
The total electric flux over a surface is obtained by<br />
summing :<br />
→ →<br />
→ →<br />
ϕ E = ∑ E . ∆ s or<br />
∫ E .d s<br />
Gauss's Theorem The total electric flux across a<br />
1<br />
closed surface is equal to times the total charge<br />
ε0<br />
inside the surface.<br />
Mathematically ∑ → E . ∆ →<br />
s = q/ε 0<br />
where q is the total charge enclosed by the surface.<br />
E<br />
N<br />
Problems in electrostatics can be greatly simplified<br />
by the use of Gaussian surfaces. These are imaginary<br />
surfaces in which the electric intensity is either<br />
parallel to or perpendicular to the surface everywhere.<br />
There are no restrictions in constructing a<br />
Gaussian surface.<br />
The following results follow from Gauss's law<br />
1. In a charged conductor, the entire charge resides<br />
only on the outer surface. (It must always be<br />
remembered that the electric field is zero inside a<br />
conductor.)<br />
2. Near a large plane conductor with a charge<br />
density σ (i.e., charge per unit area), the electric<br />
intensity is<br />
E = σ/ε 0 along the normal to the plane<br />
3. Near an infinite plane sheet of charge with a<br />
charge density σ, the electric intensity is<br />
E = σ/2ε 0 along the normal to the plane<br />
4. The electric intensity at a distance r from the axis<br />
of a long cylinder with λ charge per unit length<br />
(called the linear density of charge), is<br />
1 λ<br />
→<br />
E = along r<br />
2πε<br />
r<br />
0<br />
Problem solving strategy: Coulomb's Law :<br />
Step 1 : The relevant concepts : Coulomb's law<br />
comes into play whenever you need to know the<br />
electric force acting between charged particles.<br />
Step 2 : The problem using the following steps :<br />
Make a drawing showing the locations of the<br />
charged particles and label each particle with its<br />
charge. This step is particularly important if more<br />
than two charged particles are present.<br />
If three or more charges are present and they do<br />
not all lie on the same line, set up an xycoordinate<br />
system.<br />
Often you will need to find the electric force on<br />
just one particle. If so, identify that particle.<br />
Step 3 : The solution as follows :<br />
For each particle that exerts a force on the particle<br />
of interest, calculate the magnitude of that force<br />
1 | q1q<br />
2 |<br />
using equation F =<br />
2<br />
4πε0 r<br />
Sketch the electric force vectors acting on the<br />
particle(s) of interest due to each of the other<br />
particles (that is, make a free-body diagram).<br />
Remember that the force exerted by particle 1 on<br />
particle 2 points from particle 2 toward particle 1<br />
if the two charges have opposite signs, but points<br />
from particle 2 directly away from particle 1 if the<br />
charges have the same sign.<br />
Calculate the total electric force on the particle(s)<br />
of interest. Remember that the electric force, like<br />
any force, is a vector. When the forces acting on a<br />
charge are caused by two or more other charges,<br />
the total force on the charge is the vector sum of<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 24 MAY 2010
the indivual forces. It's often helpful to use<br />
components in an xy-coordinate system. Be sure<br />
to use correct vector notation; if a symbol<br />
represents a vector quantity, put an arrow over it.<br />
If you get sloppy with your notation, you will also<br />
get sloppy with your thinking.<br />
As always, using consistent units is essential.<br />
With the value of k = 1/4πε 0 given above,<br />
distances must be in meters, charge in coulombs,<br />
and force in newtons. If you are given distance in<br />
centimeters, inches, or furlongs, donot forget to<br />
convert ! When a charge is given in<br />
microcoulombs (µC) or nanocoulombs (nC),<br />
remember that 1µC = 10 –6 C and 1nC = 10 –9 C.<br />
Some example and problems in this and later<br />
chapters involve a continuous distribution of<br />
charge along a line or over a surface. In these<br />
cases the vector sum described in Step 3 becomes<br />
a vector integral, usually carried out by use of<br />
components. We divide the total charge<br />
distribution into infinitesimal pieces, use<br />
Coulomb's law for each piece, and then integrate<br />
to find the vector sum. Sometimes this process<br />
can be done without explicit use of integration.<br />
In many situations the charge distribution will be<br />
symmetrical. For example, you might be asked to<br />
find the force on a charge Q in the presence of<br />
two other identical charges q, one above and to<br />
the left of Q and the other below and to the left of<br />
Q. If the distance from Q to each of the other<br />
charges are the same, the force on Q from each<br />
charge has the same magnitude; if each force<br />
vector makes the same angle with the horizontal<br />
axis, adding these vectors to find the net force is<br />
particularly easy. Whenever possible, exploit any<br />
symmetries to simplify the problem-solving<br />
process.<br />
Step 4 : your answer : Check whether your numerical<br />
results are reasonable, and confirm that the like<br />
charges repel opposite charges attract.<br />
Problem solving strategy : Electric-field calculations<br />
Step 1: the relevant concepts : Use the principle of<br />
superposition whenever you need to calculate the<br />
electric field due to a charge distribution (two or<br />
more point charges, a distribution over a line, surface,<br />
or volume or a combination of these).<br />
Step 2: The problem using the following steps :<br />
Make a drawing that clearly shows the locations<br />
of the charges and your choice of coordinate axes.<br />
On your drawing, indicate the position of the field<br />
point (the point at which you want to calculate the<br />
electric field E r ). Sometimes the field point will<br />
be at some arbitrary position along a line. For<br />
example, you may be asked to find E r at point on<br />
the x-axis.<br />
Step 3 : The solution as follows :<br />
Be sure to use a consistent set of units. Distances<br />
must be in meters and charge must be in<br />
coulombs. If you are given centimeters or<br />
nanocoulombs, do not forget to convert.<br />
When adding up the electric fields caused by<br />
different parts of the charge distribution,<br />
remember that electric field is a vector, so you<br />
must use vector addition. Don't simply add<br />
together the magnitude of the individual fields:<br />
the directions are important, too.<br />
Take advantage of any symmetries in the charge<br />
distribution. For example, if a positive charge and<br />
a negative charge of equal magnitude are placed<br />
symmetrically with respect to the field point, they<br />
produce electric fields of the same magnitude but<br />
with mirror-image directions. Exploiting these<br />
symmetries will simplify your calculations.<br />
Must often you will use components to compute<br />
vector sums. Use proper vector notation;<br />
distinguish carefully between scalars, vectors, and<br />
components of vectors. Be certain the<br />
components are consistent with your choice of<br />
coordinate axes.<br />
In working out the directions of E r<br />
vectors, be<br />
careful to distinguish between the source point<br />
and the field point. The field produced by a point<br />
charge always points from source point to field<br />
point if the charge is positive; it points in the<br />
opposite direction if the charge is negative.<br />
In some situations you will have a continuous<br />
distribution of charge along a line, over a surface,<br />
or through a volume. Then you must define a<br />
small element of charge that can be considered as<br />
a point, finds of all charge elements. Usually it is<br />
easiest to do this for each component of E r<br />
separately, and often you will need to evaluate<br />
one or more integrals. Make certain the limits on<br />
your integrals are correct; especially when the<br />
situation has symmetry, make sure you don't<br />
count the charge twice.<br />
Step 4 : your answer : Check that the direction of E r<br />
is reason able. If your result for the electric-field<br />
magnitude E is a function of position (say, the<br />
coordinate x), check your result in any limits for<br />
which you know what the magnitude should be.<br />
When possible, check your answer by calculating it<br />
in a different way.<br />
Problem solving strategy : Gauss's Law<br />
Step 1 : Identify the relevant concepts : Gauss's law<br />
is most useful in situations where the charge<br />
distribution has spherical or cylindrical symmetry or<br />
is distributed uniform over a plane. In these situations<br />
we determine the direction of E r from the symmetry<br />
of the charge distribution. If we are given the charge<br />
distribution. we can use Gauss's law to find the the<br />
magnitude of E r . Alternatively, if we are given the<br />
field, we can use Gauss's law to determine the details<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 25 MAY 2010
of the charge distribution. In either case, begin your<br />
analysis by asking the question, "What is the<br />
symmetry ?"<br />
Step 2 : Set up the problem using the following steps<br />
Select the surface that you will use with Gauss's<br />
law. We often call it a Gaussian surface. If you<br />
are trying to find the field at a particular point,<br />
then that point must lie on your Gaussian surface.<br />
The Gaussian surface does not have to be a real<br />
physical surface, such as a surface of a solid<br />
body. Often the appropriate surface is an<br />
imaginary geometric surface; it may be in empty<br />
space, embedded in a solid body, or both.<br />
Usually you can evaluate the integral in Gauss's<br />
law (without using a computer) only if the<br />
Gaussian surface and the charge distribution have<br />
some symmetry property. If the charge<br />
distribution has cylindrical or spherical<br />
symmetry, choose the Gaussian surface to be a<br />
coaxial cylinder or a concentric sphere,<br />
respectively.<br />
Step 3 : Execute the solution as follows :<br />
Carry out the integral in Eq.<br />
ΦE =<br />
∫<br />
E cosφ dA =<br />
∫<br />
E dA =<br />
∫<br />
E<br />
r .dA<br />
r Q<br />
= encl<br />
ε0<br />
(various forms of Gauss's law)<br />
This may look like a daunting task, but the<br />
symmetry of the charge distribution and your<br />
careful choice of a Gaussian surface makes it<br />
straightforward.<br />
Often you can think of the closed surface as being<br />
made up of several separate surfaces, such as the<br />
side and ends of a cylinder. The integral<br />
∫ E dA<br />
over the entire closed surface is always equal to<br />
the sum of the integrals over all the separate<br />
surfaces. Some of these integrals may be zero, as<br />
in points 4 and 5 below.<br />
If E r is perpendicular (normal) at every point to a<br />
surface with area A, if points outward from the<br />
interior of the surface, and if it equal to EA. If<br />
instead E r is perpendicular and inward, then E ⊥ =<br />
– E and<br />
∫ E ⊥dA = – EA.<br />
If E r is tangent to a surface at every point, then E ⊥<br />
= 0 and the integral over that surface is zero.<br />
If E r = 0 at every point on a surface, the integral<br />
is zero.<br />
In the integral<br />
∫ E ⊥dA , E ⊥ is always the<br />
perpendicular component of the total electric field<br />
at each point on the closed Gaussian surface. In<br />
general, this field may be caused partly by<br />
charges within the surface and partly by charges<br />
outside it. Even when there is no charge within<br />
the surface, the field at points on the Gaussian<br />
surface is not necessarily zero. In that case,<br />
however, the integral over the Gaussian surface –<br />
is always zero.<br />
Once you have evaluated the integral, use eq. to<br />
solve for your target variable.<br />
Step 4 : Evaluate your answer : Often your result will<br />
be a function that describes how the magnitude of the<br />
electric field varies with position. Examine this function<br />
with a critical eye to see whether it make sense.<br />
1. Supposing that the earth has a charge surface density<br />
of 1 electron/metre 2 , calculate (i) earth's potential, (ii)<br />
electric field just outside earths surface. The<br />
electronic charge is – 1.6 × 10 –19 coulomb and earth's<br />
radius is 6.4×10 6 metre (ε 0 = 8.9 × 10 –12 coul 2 /nt–m 2 ).<br />
Sol. Let R and σ be the radius and charge surface density<br />
of earth respectively. The total charge, q on the earth<br />
surface is given by<br />
q = 4 p R 2 σ<br />
(i) The potential V at a point on earth's surface is same<br />
as if the entire charge q were concentrated at its<br />
centre. Thus,<br />
1 q<br />
V = .<br />
4πε 0 R<br />
1 4πR<br />
2 σ R. σ<br />
= . =<br />
4πε0<br />
R ε0<br />
Substituting the given values<br />
−6<br />
−19<br />
2<br />
(6.4×<br />
10 metre) × ( −1.6×<br />
10 coul / metre )<br />
V =<br />
−12<br />
2 2<br />
(8.9×<br />
10 coul / nt − m )<br />
(ii) E =<br />
Solved Examples<br />
nt − m<br />
= – 0.115 coul<br />
1 q 1<br />
2 =<br />
4πε0<br />
R 4πε . 4πR<br />
σ<br />
2<br />
0 R<br />
−19<br />
joule<br />
= – 0.115 = – 0.115 volt.<br />
coul<br />
−1.6×<br />
10 coul / metre<br />
=<br />
= – 1.8 × 10 –8 nt/coul.<br />
−12<br />
2 2<br />
8.9×<br />
10 coul / nt − m<br />
The negative sign shows that E is radially inward.<br />
2. Determine the electric field strength vector if the<br />
potential of this field depends on x, y co-ordinates as<br />
(a) V = a(x 2 – y 2 ) and (b) V = axy.<br />
Sol. (a) V = a(x 2 – y 2 )<br />
∂V<br />
∂V<br />
Hence, E x = – = – 2ax, E y = – = + 2ay<br />
∂ x ∂ y<br />
∴ E = – 2axi + 2ayj<br />
or E = – 2a(xi – yj)<br />
(b) V = a x y<br />
2<br />
2<br />
=<br />
σ<br />
ε 0<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 26 MAY 2010
Hence, E x = –<br />
∴<br />
∂V<br />
∂ x<br />
E = – ayi – axj<br />
= – a[yi + xj]<br />
∂V<br />
= –ay, E y = –<br />
∂ y<br />
= – ax<br />
3. A charge Q is distributed over two concentric hollow<br />
spheres of radii r and R (> r) such the surface<br />
densities are equal. Find the potential at the common<br />
centre.<br />
q′ q<br />
O<br />
R<br />
Sol. Let q and q′ be the charges on inner and outer sphere.<br />
Then<br />
q + q′ = Q<br />
…(1)<br />
As the surface densities are equal, hence<br />
q q'<br />
=<br />
2 2<br />
4πr<br />
4πR<br />
(∴ Surface density = charge/area)<br />
∴ q R 2 = q′ r 2<br />
…(2)<br />
From eq. (1) q′ = (Q – q), hence<br />
q R 2 = (Q – q)r 2<br />
q(R 2 + r 2 ) = Q r 2<br />
2<br />
2<br />
Q r<br />
Q R<br />
∴ q = and q′ = Q – q =<br />
2 2<br />
2 2<br />
R + r<br />
R + r<br />
Now potential at O is given by<br />
1 q 1 q'<br />
V = +<br />
4πε0<br />
r 4πε0<br />
r<br />
=<br />
=<br />
1<br />
4πε<br />
0<br />
0<br />
2<br />
Q r 1<br />
+<br />
2 2<br />
(R + r ) r 4πε<br />
Q (r + R)<br />
2<br />
4πε (R r )<br />
2 +<br />
r<br />
0<br />
(R<br />
Q r<br />
2<br />
2<br />
2<br />
+ r ) r<br />
4. S 1 and S 2 are two parallel concentric spheres enclosing<br />
charges q and 2q respectively as shown in fig.<br />
(a) What is the ratio of electric flux through S 1 and S 2 ?<br />
(b) How will the electric flux through the sphere S 1<br />
change, if a medium of dielectric constant 5 is<br />
introduced in the space inside S 1 in place of air ?<br />
S 2<br />
Sol. (a) Let Φ 1 and Φ 2 be the electric flux through spheres<br />
S 1 and S 2 respectively.<br />
q<br />
2q<br />
S 1<br />
∴<br />
q<br />
Φ 1 =<br />
ε<br />
0<br />
q / ε<br />
Φ 1 =<br />
0<br />
Φ 2 3q / ε0<br />
and Φ 2 =<br />
= 3<br />
1<br />
q + 2q 3q<br />
=<br />
ε ε<br />
(b) Let E be the electric field intensity on the surface of<br />
sphere S 1 due to charge q placed inside the sphere.<br />
When dielectric medium of dielectric constant K is<br />
introduced inside sphere S 1 , then electric field<br />
intensity E′ is given by<br />
E′ = E/K<br />
Now the flux Φ′ through S 1 becomes<br />
' 1 q<br />
Φ′ =<br />
∫<br />
E .dS =<br />
∫<br />
E.dS =<br />
K Kε<br />
∴ Φ′ =<br />
q<br />
5ε<br />
0<br />
5. A charge of 4 × 10 –8 C is distributed uniformly on the<br />
surface of a sphere of radius 1 cm. It is covered by a<br />
concentric, hollow conducting sphere of a radius<br />
5 cm. (a) Find the electric field at a point 2 cm away<br />
from the centre. (b) A charge of 6 × 10 –8 C is placed<br />
on the hollow sphere. Find the surface charge density<br />
on the outer surface of the hollow sphere.<br />
Sol. (a) See fig. (a) Let P be a point where we have to<br />
calculate the electric field. We draw a Gaussian<br />
surface (shown dotted) through point P. The flux<br />
through this surface is<br />
q = 6 × 10 –8 C<br />
5cm<br />
Φ =<br />
2cm<br />
P<br />
Fig. (a) Fig. (b)<br />
∫<br />
∫<br />
E.dS = E dS = 4π(2×<br />
10 ) E<br />
0<br />
0<br />
0<br />
−2<br />
2<br />
According to Gauss's law, Φ = q/ε 0<br />
∴ 4π × (2 × 10 –2 ) 2 E = q/ε 0<br />
9<br />
−8<br />
q (9×<br />
10 ) × (4×<br />
10 )<br />
or E =<br />
=<br />
−2<br />
2<br />
−4<br />
4πε0 × (2×<br />
10 ) 4×<br />
10<br />
= 9 × 10 5 N/C<br />
(b) See fig. (b) We draw a Gaussian surface (shown<br />
dotted) through the material of hollow sphere. We<br />
know that the electric field in a conducting material is<br />
zero, therefore the flux through this Gaussian surface<br />
is zero. Using Gauss's law, the total charge enclosed<br />
must be zero. So, the charge on the inner surface of<br />
hollow sphere is 6 × 10 –8 C. So, the charge on the<br />
outer surface will be 10 × 10 –8 C.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 27 MAY 2010
PHYSICS FUNDAMENTAL FOR <strong>IIT</strong>-<strong>JEE</strong><br />
1-D Motion, Projectile Motion<br />
KEY CONCEPTS & PROBLEM SOLVING STRATEGY<br />
Kinematics :<br />
Velocity (in a particular direction)<br />
=<br />
Displacement(in that direction)<br />
Time taken<br />
( V r AB ) x = V r Ax – V r Bx and ( V r AB ) x =<br />
dx<br />
t<br />
Where dx is the displacement in the x direction in<br />
time t.<br />
Swimmer crossing a river<br />
d<br />
v s<br />
θ<br />
v s cosθ<br />
v s sinθ<br />
d<br />
Time taken to cross the river =<br />
Vs<br />
cosθ<br />
For minimum time, θ should be zero.<br />
x<br />
v s<br />
v r<br />
in this case resultant velocity<br />
2 2 d<br />
V R = V s + vr<br />
and t = .<br />
v s<br />
Also x = v r × t<br />
For reaching a point just opposite the horizontal<br />
component of velocity should be zero.<br />
v sinθ = V r<br />
| Displacement |<br />
|Average Velocity| =<br />
time<br />
a r v<br />
r – u<br />
r r r<br />
=<br />
t<br />
= v + (–u)<br />
t<br />
2<br />
2<br />
v + u – 2uvcosθ<br />
⇒ |a| =<br />
t<br />
Where θ is the angle between v and u.<br />
The direction of acceleration is along the resultant of<br />
v r and (– u r ).<br />
v r<br />
v R<br />
Graphs<br />
During analysis of a graph, the first thing is see the<br />
physical quantities drawn along x-axis and y-axis.<br />
If y = mx, the graph is a straight line passing through<br />
the origin with slope = m. [see fig. (a)]<br />
Y<br />
m = tanθ<br />
θ is acute and<br />
m is positive<br />
Y<br />
m = tanθ<br />
θ is abtuse and<br />
m is positive<br />
θ<br />
θ<br />
X<br />
X<br />
(i) fig.(a) (ii)<br />
if y = mx + c, the graph is a straight line not passing<br />
through the origin and having an intercept c which<br />
may be positive or negative [see fig. (b,) (c)]]<br />
Y<br />
Y<br />
c is negative<br />
m is '+' ve<br />
m is '+' ve<br />
c<br />
(i)<br />
Y<br />
X<br />
fig(b)<br />
c<br />
(ii)<br />
c is positive<br />
m is negative<br />
(ii) X<br />
fig(c)<br />
For y = kx 2 , where k is a constant, we get parabola<br />
[see fig (d)]<br />
Y<br />
Parabola<br />
Fig.(d)<br />
x 2 + y 2 = r 2 is equation of a circle with centre at<br />
origin and radius r.<br />
X<br />
X<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 28 MAY 2010
For (x – a) 2 + (y – b) 2 = r 2 , the motion is in a circular<br />
path with centre at (a, b) and radius r<br />
2<br />
2<br />
x y<br />
2 +<br />
2 = 1 is equation of an ellipse<br />
a b<br />
x × y = constant give a rectangular hyperbola.<br />
Note : To decide the path of motion of a body, a<br />
relationship between x and y is required.<br />
Area under-t graph represents change in velocity.<br />
Calculus method is used for all types of motion (a = 0<br />
or a = constt or a = variable)<br />
s = f(t)<br />
Differentiate w.r.t<br />
time<br />
v = f(t)<br />
integrate w.r.t.<br />
time<br />
Differentiate w.r.t<br />
time<br />
a = f(t)<br />
integrate w.r.t.<br />
time<br />
S stand for displacement<br />
2<br />
dv dv d s<br />
a = v = = ds dt 2<br />
dt<br />
dx dy<br />
Also v x = ⇒ vy = dt dt<br />
2<br />
dv<br />
a x = x d x dv 2 y d y<br />
= and ay = =<br />
dt 2<br />
dt<br />
dt 2<br />
dt<br />
The same concept can be applied for z-co-ordintae.<br />
Projectile motion :<br />
P<br />
ucosθ<br />
u ucosθ<br />
usinθ ucosθ<br />
θ<br />
ucosθ<br />
ucosθ<br />
Q θ<br />
u<br />
usinθ<br />
g<br />
g F<br />
F<br />
g<br />
F<br />
g<br />
F<br />
Projectile motion is a uniformly accelerated motion.<br />
For a projectile motion, the horizontal component of<br />
velocity does not change during the path because<br />
there is no force in the horizontal direction. The<br />
vertical component of velocity goes on decreasing<br />
with time from O to P. At he highest point it becomes<br />
zero. From P to Q again. the vertical component of<br />
velocity increases but in downwards direction.<br />
Therefore the minimum velocity is at the topmost<br />
point and it is u cos θ directed in the horizontal<br />
direction.<br />
The mechanical energy of a projectile remain<br />
constant throughout the path.<br />
the following approach should be adopted for solving<br />
problems in two-dimensional motion :<br />
Resolve the 2-D motion in two 1-D motions in two<br />
mutually perpendicular directions (x and y direction)<br />
Resolve the vector quantitative along these<br />
directions. Now use equations of motion separately<br />
for x-direction and y-directions.<br />
If you do not resolve a 2-D motions in two 1-D<br />
motions in two 1-D motion then use equations of<br />
motion in vector form<br />
v r = u r + at ; s r 1 r<br />
= ut + a t<br />
2<br />
; v r . v r – u r . u = 2 a r s r<br />
2<br />
s = 2<br />
1 ( u<br />
r + v<br />
r )t<br />
When y = f(x) and we are interested to find<br />
(a) The values of x for which y is maximum for<br />
minimum<br />
(b) The maximum/minimum values of y then we may<br />
use the concept of maxima and minima.<br />
Problem solving strategy :<br />
Motion with constant Acceleration :<br />
Step 1: Identify the relevant concepts : In most<br />
straight-line motion problems, you can use the<br />
constant-acceleration equations. Occasionally,<br />
however, you will encounter a situation in which the<br />
acceleration isn't constant. In such a case, you'll need<br />
a different approach<br />
dυ<br />
a x = x d ⎛ dx ⎞ d x<br />
= ⎜ ⎟ =<br />
dt dt<br />
2<br />
⎝ dt ⎠ dt<br />
Step 2: Set up the problem using the following steps:<br />
You must decide at the beginning of a problem<br />
where the origin of coordinates are usually a<br />
matter of convenience. If is often easiest to place<br />
the particle at the origin at time t = 0; then x 0 = 0.<br />
It is always helpful to make a motion diagram<br />
showing these choices and some later positions of<br />
the particle.<br />
Remember that your choice of the positive axis<br />
direction automatically determines the positive<br />
directions for velocity and acceleration. If x is<br />
positive to the right of the origin, the v x and a x are<br />
also positive toward the right.<br />
Restate the problem in words first, and then<br />
translate this description into symbols and<br />
equations. When does the particle arrive at a<br />
certain point (that is, what is the value of t)?<br />
where is the particle when its velocity has a<br />
specified value (that is, what is the value of x<br />
when v x has the specified value)? "where is the<br />
motorcyclist when his velocity is 25m/s?"<br />
2<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 29 MAY 2010
Translated into symbols, this becomes "What is<br />
the value of x when v x = 25 m/s?"<br />
Make a list of quantities such as x, x 0 ,v x ,v 0x ,a x and<br />
t. In general, some of the them will be known<br />
quantities, and decide which of the unknowns are<br />
the target variables. Be on the lookout for implicit<br />
information. For example. "A are sits at a<br />
stoplight" Usually means v 0x = 0.<br />
Step 3 : Execute the solution :<br />
Choose an equation from Equation v x = v 0x + a x t<br />
x = x 0 + v 0x t + 2<br />
1<br />
ax t 2<br />
2<br />
v x =<br />
(constant acceleration only)<br />
2<br />
v 0x<br />
+ 2a x (x – x 0 ) (constant accelerations only)<br />
⎛ v0x<br />
+ vx<br />
⎞<br />
x – x 0 = ⎜ ⎟ t (constant acceleration only)<br />
⎝ 2 ⎠<br />
that contains only one of the target variables. Solve<br />
this equation for the equation for the target variable,<br />
using symbols only. then substitute the known values<br />
and compute the value of the target variable.<br />
sometimes you will have to solve two simultaneous<br />
equations for two unknown quantities.<br />
Step 4 : Evaluate your answer : Take a herd look at<br />
your results to see whether they make sense. Are<br />
they within the general range of values you<br />
expected?<br />
Problem solving strategy :<br />
Projectile Motion :<br />
Step 1 : Identify the relevant concepts : The key<br />
concept to remember is the throughout projectile<br />
motion, the acceleration is downward and has a<br />
constant magnitude g. Be on the lookout for aspects<br />
of the problem that do not involve projectile motion.<br />
For example, the projectile-motion equations don't<br />
apply to throwing a ball, because during the throw<br />
the ball is acted on by both the thrower's hand and<br />
gravity. These equations come into play only after<br />
the ball leaves the thrower's hand.<br />
Step 2 : Set up the problem using the following steps<br />
Define your coordinate system and make a sketch<br />
showing axes. Usually it's easiest to place the<br />
origin to place the origin at the initial (t = 0)<br />
position of the projectile. (If the projectile is a<br />
thrown ball or a dart shot from a gun, the<br />
thrower's hand or exits the muzzle of the gun.)<br />
Also, it's usually best to take the x-axis as being<br />
horizontal and the y-axis as being upward. Then<br />
the initial position is x 0 = 0 and y 0 = 0, and the<br />
components of the (constant) acceleration are a x = 0,<br />
a y = – g.<br />
List the unknown and known quantities, and<br />
decide which unknowns are your target variables.<br />
In some problems you'll be given the initial<br />
velocity (either in terms of components or in<br />
terms of magnitude and direction) and asked to<br />
find the coordinates and velocity components as<br />
some later time. In other problems you might be<br />
given two points on the trajectory and asked to<br />
find the initial velocity. In any case, you'll be<br />
using equations<br />
x = (v 0 cosα 0 )t (projectile motion) through ...(1)<br />
v y = v 0 sin α 0 – gt (projectile motion) ...(2)<br />
make sure that you have as many equations as<br />
there are target variables to be found.<br />
It often helps to state the problem in words and<br />
then translate those words into symbols. For<br />
example, when does the particle arrive at a certain<br />
point ? (That is at what value of t?) Where is the<br />
particle when its velocity has a certain value?<br />
(That is, what are the values of x and y when v x or<br />
v y has the specified value ?) At the highest point<br />
in a trajectory, v y = 0. so the question "When does<br />
the particle reach its highest points ?" translates<br />
into "When does the projectile return to its initial<br />
elevation?" translates into "What is the value of t<br />
when y = y 0 ?"<br />
Step 3 : Execute the solution use equation (1) & (2)<br />
to find the target variables. As you do so, resist the<br />
temptation to break the trajectory into segments and<br />
analyze each segment separately. You don't have to<br />
start all over, with a new axis and a new time scale,<br />
when the projectile reaches its highest point ! It's<br />
almost always easier to set up equation (1) & (2)<br />
at the starts and continue to use the same axes and<br />
time scale throughout the problem.<br />
Step 4 : Evaluate your answer : As always, look at<br />
your results to see whether they make sense and<br />
whether the numerical values seem reasonable.<br />
Relative Velocity :<br />
Step 1 : Identify the relevant concepts : Whenever<br />
you see the phrase "velocity relative to" or "velocity<br />
with respect to", it's likely that the concepts of<br />
relative will be helpful.<br />
Step 2 : Set up the problem : Label each frame of<br />
reference in the problem. Each moving object has its<br />
own frame of reference; in addition, you'll almost<br />
always have to include the frame of reference of the<br />
earth's surface. (Statements such as "The car is<br />
traveling north at 90 km/h" implicitly refer to the<br />
car's velocity relative to the surface of the earth.) Use<br />
the labels to help identify the target variable. For<br />
example, if you want to find the velocity of a car (C)<br />
with respect to a bus (B), your target variable is v C/B .<br />
Step 3 : Execute the solution : Solve for the target<br />
variable using equation<br />
v P/A = v P/B + v B/A (relative velocity along a line) ...(1)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 30 MAY 2010
(If the velocities are not along the same direction, (As the ball returns to its initial position, the change<br />
→<br />
(d) horizontal range of the projectile<br />
→<br />
Since s 1 = s 2 = d and s net = s + s | = 0<br />
you'll need to use the vector from of this equations,<br />
derived later in this section.) It's important to note the<br />
in position, the change in position vector of the ball,<br />
that is the net displacement will be zero).<br />
order of the double subscripts in equation (1) v A/B<br />
→<br />
always means "velocity of A relative to B." These ∴ | VaV<br />
| = 0.<br />
subscripts obey an interesting kind of algebra, as<br />
equation (1) shown. If regard each one as a fraction,<br />
2. A long belt is moving horizontally with a speed of 4<br />
then the fraction on the left side is the product of the<br />
Km/hour. A child runs on this belt to and fro with a<br />
fractions on the right sides : P/A = (P/B) (B/A). This<br />
speed of 9 Km/hour (with respect to the belt) between<br />
is a handy rule you can use when applying Equation<br />
his father and mother located 50 m apart on the<br />
(1) to any number of frames of reference. For<br />
moving belt. For an observer on a stationary platform<br />
example, if there are three different frames of<br />
outside, what is the<br />
reference A, B, and C, we can write immediately. (a) speed of the child running in the direction of motion<br />
v P/A = v P/C + v C/B + V B/A<br />
of the belt,<br />
Step 4 : Evaluate your answer : Be on the lookout for<br />
stray minus signs in your answer. If the target<br />
(b) speed of the child running opposite to the direction of<br />
motion of the belt and<br />
variable is the velocity of a car relative to a bus<br />
(v V/B ), make sure that you haven't accidentally<br />
calculated the velocity of the bus relative of the car<br />
(c) time taken by the child in case (a) and (b) ?<br />
Which of the answers change, if motion is viewed by<br />
one of the parents ?<br />
(v B/C ). If you have made this mistake, you can<br />
recover using equation.<br />
v A/B = – v B/A<br />
Sol. Let us consider positive direction of x-axis from left<br />
to right<br />
(a) Here, v B = + 4 Km/hour<br />
Speed of child w.r.t. belt, v C = = 9 Km/hour<br />
Solved Examples<br />
∴ Speed of child w.r.t. stationary observer,<br />
v C ′ = v C + v B or v C ′ = 9 + 4 = 13 Km/hour<br />
1. A small glass ball is pushed with a speed V from A.<br />
It moves on a smooth surface and collides with the<br />
wall at B. If it loses half of its speed during the<br />
collision, find the distance, average speed and<br />
velocity of the ball till it reaches at its initial position.<br />
(b) Here, v B = + 4 Km/hour, v C = – 9 Km/hour<br />
∴ Speed of child w.r.t. stationary observer,<br />
v C ′ = v C + v B or v C ′ = – 9 + 4 = –5 Km/hour<br />
The negative sign shows that the child appears to run<br />
in a direction opposite to the direction of motion of<br />
the belt.<br />
A V 0.5V B<br />
(c) Distance between the parents, s = 50 m = 0.05 Km<br />
Since parents and child are located on the same belt,<br />
the speed of the child as observe by stationary<br />
d<br />
Sol. The ball moves from A to B with a constant speed V.<br />
Since it loses half of its speed on collision, it returns<br />
observer in either direction (either father to mother or<br />
from mother to father) will be 9 Km/hour.<br />
Time taken by the child in case (a) and (b),<br />
from B to A with a constant speed V/2.<br />
0.50 km<br />
∴ V 1 = V and V 2 = V/2<br />
t = = 20 sec.<br />
9 km / hour<br />
d1<br />
+ d 2<br />
Using the formula, V aV =<br />
If the motion is observed by one of parents, answer to<br />
(d1<br />
/ V 1)<br />
+ (d 2 / V2<br />
)<br />
case (a) case (b) gets altred. It is because the speed<br />
Putting d 1 = d 2 = d; V 1 = V and V 2 = V/2<br />
of the child w.r.t. either of mother or father is<br />
d1<br />
+ d 2 2V<br />
9 Km/hour.<br />
We obtain, V aV =<br />
=<br />
(dV) + (d / 0.5V) 3<br />
3. A particle is projected with velocity v<br />
From the formula,<br />
0 = 100 m/s at<br />
an angle θ = 30º with the horizontal. Find :<br />
→ → →<br />
→<br />
| s1+<br />
s2<br />
| | snet<br />
| (a) velocity of the particle after 2 sec.<br />
average velocity = VaV<br />
= =<br />
s1<br />
s2<br />
t<br />
(b) angle between initial velocity and the velocity after 2 sec.<br />
+<br />
net<br />
V1<br />
V2<br />
(c) the maximum height reached by the projectile<br />
| 1 2<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 31 MAY 2010
Sol. (a)<br />
(b)<br />
(c)<br />
→<br />
v<br />
t<br />
→<br />
= v<br />
xt<br />
→<br />
î + v<br />
yt<br />
ĵ<br />
where î and ĵ are the unit vectors along +ve x and<br />
+ve y-axis respectively<br />
→<br />
t<br />
v =(u x + a x t) î + (u y + a y t) ĵ<br />
Here, u x = v 0 cos θ = 50 3 m/s, a x = 0<br />
u y = v 0 sin θ = 50 m/s, a y = – g<br />
(Q g acts downwards)<br />
→<br />
t<br />
v = 50<br />
3 î + (50 – 10 × 2) ĵ<br />
=[50 3 î + 30 ĵ ] m/s<br />
∴ | → 2 2<br />
v 2 | = (vx + v y ) = 2 2<br />
( 50 3) + (30)<br />
∴<br />
→<br />
0<br />
→<br />
2<br />
→<br />
0<br />
v = 50<br />
v = 50<br />
3 î + 50 ĵ<br />
3 î + 30 ĵ<br />
v . v → 2 = 7500 + 1500 = 9000<br />
If α is the angle between v → 0 and v<br />
→ 2<br />
→<br />
v0<br />
.v2<br />
9000<br />
Then, cos α = =<br />
→ →<br />
100×<br />
91.65<br />
| v0<br />
| × | v2<br />
|<br />
α = cos –1 (0.98) = 10.8º<br />
v 2 y – u 2 y = 2a y y<br />
At y = y max , v y = 0<br />
∴ 0 – v 2 0 sin 2 θ = 2 (–g)y max<br />
2 2<br />
v0 sin θ<br />
∴ y max = = 125 m<br />
2g<br />
→<br />
u 2 sin 2θ (d) R = = 1732 m<br />
g<br />
4. A ball starts falling with zero initial velocity on a<br />
smooth inclined plane forming an angle α with the<br />
horizontal. Having fallen the distance 'h', the ball<br />
rebounds elastically off the inclined plane. At what<br />
distance from the impact point will the ball rebound<br />
for the second time ?<br />
α<br />
Sol. Just before impact magnitude of velocity of the ball,<br />
v = ( 2gh)<br />
α<br />
α<br />
As the ball collides elastically and the inclined plane<br />
is fixed, the ball follows the law of reflection.<br />
Now along the incline, velocity component after<br />
impact is v sin α and acceleration is g sin α.<br />
Perpendicular to the incline, velocity component is<br />
vcos α and acceleration (– g cos α). Hence, if we<br />
measure x and y-coordinate along the incline and<br />
perpendicular to the incline, then<br />
x = (v sin α) t + ½ (g sin α)t 2<br />
and y = (v cos α) t – ½ (g cos α)t 2<br />
When the ball hits the plane for a second time,<br />
y = 0, (v cos α)t – ½(g cos α)t 2 or t = (2v/g)<br />
Putting this value of t in x,<br />
4v<br />
2 sin α<br />
x = = 8h sin α<br />
g<br />
5. A batsman hits a ball at a height of 1.22m above the<br />
ground so that ball leaves the bat at an angle 45º with<br />
the horizontal. A 7.31 m high wall is situated at a<br />
distance of 97.53 m from the position of the batsman.<br />
Will the ball clear the wall if its range is 106.68 m.<br />
Take g = 10 m/s 2<br />
v0<br />
2 sin 2θ<br />
Sol. R(range) =<br />
g<br />
or,<br />
1.22m<br />
2 Rg<br />
v 0 = = Rg as θ = 45º<br />
sin 2θ<br />
A<br />
45º<br />
v 0<br />
106.68m<br />
or, v 0 = ( Rg)<br />
…(1)<br />
Equation of trajectory<br />
or, y = x –<br />
y = x tan 45º –<br />
2v<br />
gx 2<br />
2Rg.½<br />
2<br />
0<br />
gx<br />
2<br />
cos<br />
2<br />
gx 2<br />
= x – Rg<br />
B<br />
45º<br />
Putting x = 97.53, we get<br />
2<br />
10×<br />
(97.53)<br />
y = 97.53 –<br />
= 8.35 cm<br />
106.68×<br />
10<br />
Hence, height of the ball from the ground level is<br />
h = 8.35 + 1.22 = 9.577 m<br />
As height of the wall is 7.31 m so the ball will clear<br />
the wall.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 32 MAY 2010
KEY CONCEPT<br />
Physical<br />
Chemistry<br />
Fundamentals<br />
GASEOUS STATE<br />
& REAL GASES<br />
Real Gases :<br />
Deviation from Ideal Behaviour :<br />
Real gases do not obey the ideal gas laws exactly<br />
under all conditions of temperature and pressure.<br />
Experiments show that at low pressures and<br />
moderately high temperatures, gases obey the laws of<br />
Boyle, Charles and Avogadro approximately, but as<br />
the pressure is increased or the temperature is<br />
decreased, a marked departure from ideal behaviour<br />
is observed.<br />
Ideal gas<br />
p<br />
V<br />
Plot of p versus V of hydrogen, as<br />
compared to that of an ideal gas<br />
The curve for the real gas has a tendency to coincide<br />
with that of an ideal gas at low pressures when the<br />
volume is large. At higher pressures, however,<br />
deviations are observed.<br />
Compressibility Factor :<br />
The deviations can be displayed more clearly, by<br />
plotting the ratio of the observed molar volume V m to<br />
the ideal molar volume V m,ideal (= RT/p) as a function<br />
of pressure at constant temperature. This ratio is<br />
called the compressibility factor Z and can be<br />
expressed as<br />
Z =<br />
V<br />
V<br />
m<br />
m,ideal<br />
p<br />
= RT<br />
Vm<br />
Plots of Compressibility Factor versus Pressure :<br />
For an ideal gas Z = 1 and is independent of pressure<br />
and temperature. For a real gas, Z = f(T, p), a<br />
function of both temperature and pressure.<br />
A graph between Z and p for some gases at 273.15 K,<br />
the pressure range in this graph is very large. It can<br />
be noted that:<br />
(1) Z is always greater than 1 for H 2 .<br />
(2) For N 2 , Z < 1 in the lower pressure range and is<br />
greater than 1 at higher pressures. It decreases with<br />
increase of pressure in the lower pressure region,<br />
passes through a minimum at some pressure and then<br />
H 2<br />
increases continuously with pressure in the higher<br />
pressure region.<br />
(3) For CO 2 , there is a large dip in the beginning. In<br />
fact, for gases which are easily liquefied, Z dips<br />
sharply below the ideal line in the low pressure<br />
region.<br />
1.0<br />
t = 0ºC<br />
H 2<br />
N 2<br />
CH 4<br />
ideal gas<br />
CO 2<br />
Z<br />
0 100 200 300<br />
p/101.325 bar<br />
Plots of Z versus p of a few gases<br />
This graph gives an impression that the nature of the<br />
deviations depend upon the nature of the gas. In fact,<br />
it is not so. The determining factor is the temperature<br />
relative to the critical temperature of the particular<br />
gas; near the critical temperature, the pV curves are<br />
like those for CO 2 , but when far away, the curves are<br />
like those for H 2 (below fig.)<br />
Z<br />
1.0<br />
T 1 >T 2 >T 3 >T 4<br />
ideal gas<br />
0 200 400 600<br />
p/101.325 kPa<br />
T 4<br />
T 3<br />
T 2<br />
Plots of Z versus p of a single gas<br />
at various temperatures<br />
Provided the pressure is of the order of 1 bar or less,<br />
and the temperature is not too near the point of<br />
liquefaction, the observed deviations from the ideal<br />
gas laws are not more than a few percent. Under<br />
these conditions, therefore, the equation pV = nRT<br />
and related expressions may be used.<br />
Van der Waals Equation of state for a Real gas<br />
Causes of Deviations from Ideal Behaviour :<br />
The ideal gas laws can be derived from the kinetic<br />
theory of gases which is based on the following two<br />
important assumptions:<br />
T 1<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 33 MAY 2010
(i) The volume occupied by the molecules is<br />
negligible in comparison to the total volume of<br />
the gas.<br />
(ii) The molecules exert no forces of attraction upon<br />
one another.<br />
Derivation of van der Waals Equation :<br />
Van der Waals was the first to introduce<br />
systematically the correction terms due to the above<br />
two invalid assumptions in the ideal gas equation<br />
p i V i = nRT. His corrections are given below.<br />
Correction for volume :<br />
V i in the ideal gas equation represents an ideal<br />
volume where the molecules can move freely. In real<br />
gases, a part of the total volume is, however,<br />
occupied by the molecules of the gas. Hence, the free<br />
volume V i is the total volume V minus the volume<br />
occupied by the molecules. If b represents the<br />
effective volume occupied by the molecules of 1<br />
mole of a gas, then for the amount n of the gas V i is<br />
given by<br />
V i = V – nb ...(1)<br />
Where b is called the excluded volume or co-volume.<br />
The numerical value of b is four times the actual<br />
volume occupied by the gas molecules. This can be<br />
shown as follows.<br />
If we consider only bimolecular collisions, then the<br />
volume occupied by the sphere of radius 2r<br />
represents the excluded volume per pair of<br />
molecules as shown in below Fig.<br />
excluded<br />
volume<br />
2r<br />
Excluded volume per pair of molecules<br />
Thus, excluded volume per pair of molecules<br />
4<br />
= π(2r) 3 ⎛ 4 ⎞<br />
= 8 ⎜ πr<br />
3 ⎟ 3 ⎝ 3 ⎠<br />
Excluded volume per molecule<br />
1 ⎡ ⎛ 4 ⎞⎤<br />
= ⎢ ⎜ π 3 ⎛ 4 ⎞<br />
8 r ⎟⎥ = 4 ⎜ πr<br />
3 ⎟<br />
2 ⎣ ⎝ 3 ⎠ ⎦ ⎝ 3 ⎠<br />
= 4 (volume occupied by a molecule)<br />
Since b represents excluded volume per mole of the<br />
gas, it is obvious that<br />
⎡ ⎛ 4 ⎞⎤<br />
b = N A ⎢4⎜<br />
πr<br />
3 ⎟⎥ ⎣ ⎝ 3 ⎠ ⎦<br />
Correction for Forces of Attraction :<br />
Consider a molecule A in the bulk of a vessel as<br />
shown in Fig. This molecule is surrounded by other<br />
molecules in a symmetrical manner, with the result<br />
that this molecule on the whole experiences no net<br />
force of attraction.<br />
A<br />
Arrangement of molecules within and<br />
near the surface of a vessel<br />
Now, consider a molecule B near the side of the<br />
vessel, which is about to strike one of its sides, thus<br />
contributing towards the total pressure of the gas.<br />
There are molecules only on one side of the vessel,<br />
i.e. towards its centre, with the result that this<br />
molecule experiences a net force of attraction<br />
towards the centre of the vessel. This results in<br />
decreasing the velocity of the molecule, and hence its<br />
momentum. Thus, the molecule does not contribute<br />
as much force as it would have, had there been no<br />
force of attraction. Thus, the pressure of a real gas<br />
would be smaller than the corresponding pressure of<br />
an ideal gas, i.e.<br />
p i = p + correction term ...(2)<br />
This correction term depends upon two factors:<br />
(i) The number of molecules per unit volume of the<br />
vessel Large this number, larger will be the net force<br />
of attraction with which the molecule B is dragged<br />
behind. This results in a greater decrease in the<br />
velocity of the molecule B and hence a greater<br />
decrease in the rate of change of momentum.<br />
Consequently, the correction term also has a large<br />
value. If n is the amount of the gas present in the<br />
volume V of the container, the number of molecules<br />
per unit volume of the container is given as<br />
nN<br />
N' = A n<br />
or N' ∝<br />
V<br />
V<br />
Thus, the correction term is given as :<br />
Correction term ∝ n/V ...( 2a)<br />
(ii) The number of molecules striking the side of the<br />
vessel per unit time Larger this number, larger will<br />
be the decrease in the rate of change of momentum.<br />
Consequently, the correction term also has a larger<br />
value,. Now, the number of molecules striking the<br />
side of vessel in a unit time also depends upon the<br />
number of molecules present in unit volume of the<br />
container, and hence in the present case:<br />
Correction term ∝ n / V<br />
...(2b)<br />
Taking both these factors together, we have<br />
B<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 34 MAY 2010
⎛ n ⎞ ⎛ n ⎞<br />
Correction term ∝ ⎜ ⎟ ⎜ ⎟<br />
⎝ V ⎠ ⎝ V ⎠<br />
2<br />
n<br />
or Correction term = a<br />
...( 3)<br />
2<br />
V<br />
Where a is the proportionality constant and is a<br />
measure of the forces of attraction between the<br />
molecules. Thus<br />
2<br />
n<br />
p i = p + a<br />
...(4)<br />
2<br />
V<br />
The unit of the term an 2 /V 2 will be the same as that of<br />
the pressure. Thus, the SI unit of a will be Pa m 6 mol –2 .<br />
It may be conveniently expressed in kPa dm 6 mol –2 .<br />
When the expressions as given by Eqs (1) and (4) are<br />
substituted in the ideal gas equation p i V i = nRT, we<br />
get<br />
⎛<br />
2<br />
⎞<br />
⎜<br />
n a<br />
p + ⎟ (V – nb) = nRT ...(5)<br />
2<br />
⎝ V ⎠<br />
This equation is applicable to real gases and is known<br />
as the van der Waals equation.<br />
Values of van der Waals Constants :<br />
The constants a and b in van der Waals equation are<br />
called van der Waals constants and their values<br />
depend upon the nature of the gas. They<br />
Van Der Waals Constants<br />
a<br />
Gas 6 2<br />
kPa dm mol –<br />
H 2<br />
He<br />
N 2<br />
O 2<br />
Cl 2<br />
NO<br />
NO 2<br />
H 2 O<br />
CH 4<br />
C 2 H 6<br />
C 3 H 8<br />
C 4 H 10 (n)<br />
C 4 H 10 (iso)<br />
C 5 H 12 (n)<br />
CO<br />
CO 2<br />
21.764<br />
3.457<br />
140.842<br />
137.802<br />
657.903<br />
135.776<br />
535.401<br />
553.639<br />
228.285<br />
556.173<br />
877.880<br />
1466.173<br />
1304.053<br />
1926.188<br />
150.468<br />
363.959<br />
b<br />
dm<br />
3 mol –1<br />
0.026 61<br />
0.023 70<br />
0.039 13<br />
0.031 83<br />
0.056 22<br />
0.027 89<br />
0.044 24<br />
0.030 49<br />
0.042 78<br />
0.063 80<br />
0.084 45<br />
0.122 6<br />
0.114 2<br />
0.146 0<br />
0.039 85<br />
0.042 67<br />
are characteristics of the gas. The values of these<br />
constants are determined by the critical constants of<br />
the gas. Actually, the so-called constant vary to some<br />
extent with temperature and this shows that the van<br />
der Waals equation is not a complete solution of the<br />
behaviour of real gases.<br />
Applicability of the Van Der Waals Equation :<br />
Since the van der Waals equation is applicable to real<br />
gases, it is worth considering how far this equation<br />
can explain the experimental behaviours of real<br />
gases. The van der Waals equation for 1 mole of a<br />
gas is<br />
⎛ ⎞<br />
⎜<br />
a<br />
p + ⎟ (V 2<br />
m – b) = RT ..(i)<br />
⎝ V m ⎠<br />
At low pressure When pressure is low, the volume is<br />
sufficiently large and b can be ignored in comparison<br />
to V m in Eq. (i). Thus, we have<br />
⎛ ⎞<br />
⎜<br />
a<br />
⎟<br />
a<br />
p + V 2<br />
m = RT or pV m + =RT<br />
⎝ V m ⎠<br />
V m<br />
a<br />
or Z = 1 –<br />
...(ii)<br />
VmRT<br />
From the above equation it is clear that in the low<br />
pressure region, Z is less than 1. On increasing the<br />
pressure in this region, the value of the term<br />
(a/V m RT) increase as V is inversely proportional to p.<br />
Consequently, Z decreases with increase of p.<br />
At high pressure When p is large , V m will be small<br />
and one cannot ignore b in comparison to V m .<br />
2<br />
However, the term a / V m may be considered<br />
negligible in comparison to p in Eq. (i) Thus,<br />
pb<br />
p(V m – b) = RT or Z = 1 + ...(iiii)<br />
RT<br />
Here Z is greater than 1 and increases linearly with<br />
pressure. This explains the nature of the graph in the<br />
high pressure region.<br />
A high temperature and low pressure If<br />
temperature is high, V m will also be sufficiently large<br />
2<br />
and thus the term a / V m will be negligibly small. At<br />
this stage, b may also be negligible in comparison to<br />
V m . Under these conditions, Eq. (i) reduces to an<br />
ideal gas equation of state:<br />
pV m = RT<br />
Hydrogen and helium The value of a is extremely<br />
small for these gases as they are difficult to liquefy.<br />
Thus, we have the equation of state as p(V m – b) = RT,<br />
obtained from the van der Waals equation by<br />
2<br />
ignoring the term a / V m . Hence, Z is always greater<br />
than 1 and it increases with increase of p.<br />
The van dar Waals equation is a distinct<br />
improvement over the ideal gas law in that it gives<br />
qualitative reasons for the deviations from ideal<br />
behaviour. However, the generality of the equation is<br />
lost as it contains two constants, the values of which<br />
depend upon the nature of the gas.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 35 MAY 2010
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 36 MAY 2010
KEY CONCEPT<br />
Organic<br />
Chemistry<br />
Fundamentals<br />
GENERAL ORGANIC<br />
CHEMISTRY<br />
Stability of different types of carbocations in<br />
decreasing order :<br />
⊕<br />
C<br />
⊕<br />
> CH<br />
⊕<br />
⊕ ⊕<br />
(Ph) 3 C > (Ph)2 CH > Ph – C H2 ≥<br />
⊕ ⊕ ⊕<br />
CH 2 = CH – CH 2 ≥ R – C – R > R – CH – R<br />
><br />
R<br />
⊕ ⊕<br />
> R – CH 2 > CH 2 = CH<br />
A special stability is associated with cycloproyl<br />
methyl cations and this stability increases with every<br />
additional cyclopropyl group.<br />
This is undoubtedly because of conjugation between<br />
the bent orbitals of the cyclopropyl ring and the<br />
vacant p-orbital of the cation carbon.<br />
H<br />
H<br />
H<br />
H<br />
Cyclopropyl methyl cation orbital representation conjugation<br />
with the p-like orbital of the ring<br />
Nucleophilicity versus basicity :<br />
If the nucleophilic atoms are from the same period of<br />
the periodic table, strength as a nucleophile parallels<br />
strength as a base. For example :<br />
H 2 O < NH 3<br />
CH 3 OH ≈ H 2 O < CH 3 CO Θ 2 < CH 3 O Θ ≈ OH Θ<br />
⇒<br />
Increasing base strength<br />
Increasing nucleophile strength<br />
Nucleophile strength increases down a column of the<br />
periodic table (in solvents that can have hydrogen<br />
bond, such as water and alcohols). For example :<br />
Θ<br />
R O <<br />
Θ<br />
R S<br />
R 3 N < R 3 P<br />
C<br />
H<br />
H<br />
⊕<br />
><br />
Θ<br />
F < Θ Cl < Θ Br < Θ I<br />
increasing nucleophilic strength<br />
decreasing base strength<br />
⇒ Steric bulk decreases nucleophilicity. For<br />
example :<br />
CH 3<br />
Θ<br />
H 3 C – C – O < HO Θ<br />
CH 3<br />
weaker nucleophile<br />
Stronger base<br />
stronger nucleophile<br />
weaker base<br />
Leaving Groups :<br />
A good leaving groups is the one which becomes a<br />
stable ion after its departure. As most leaving groups<br />
leave as a negative ion, the good leaving groups are<br />
those ions which stabilize this negative charge most<br />
effectively. The weak bases do this best, thus the best<br />
groups are weak bases. If a group is a weak base i.e.,<br />
the conjugate base of a strong acid, it will generally<br />
be a good leaving group. In an S N 2 reaction the<br />
leaving group begins to gain negative charge as the<br />
transition state is reached. The more the negative<br />
charge is stabilized, the lower is the energy of the<br />
transition state; this lowers the energy of activation<br />
and thereby increases the rate of reaction.<br />
The acids HCl, HBr, HI and H 2 SO 4 are all strong<br />
acids since the anions Cl – , Br – , I – and HSO – 4 are<br />
stable anions these anions (weak bases) are also good<br />
leaving groups in S N 2 reactions. Of the halogens, an<br />
iodide ion is the best leaving group and the fluoride<br />
ion is the poorest :<br />
I – > Br – > Cl – > F –<br />
The order of basicity is opposite : F – > Cl – > Br – > I – ,<br />
the reason that alkyl fluorides are ineffective<br />
substrates in S N 2 reactions is related, to the relatively<br />
low acidity of HF (pK a = 3). Sulfonic acids, R<br />
SO 2 OH are similar to sulfuric acid in acidity and the<br />
sulfonate ion RSO – 3 is a very good leaving group.<br />
Alky benzenesulfonates, alkyl p-toluenesulfonates<br />
are therefore, very good substrates in S N 2 reactions.<br />
The triflate ion (CF 3 SO – 3 ) is one of the best leaving<br />
groups known, it is the anion of CF 3 SO 3 H which is a<br />
strong acid much stronger than sulfuric acid.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 37 MAY 2010
Kinetic Isotope Effects :<br />
The kinetic isotope effect is a change of rate that<br />
occurs upon isotopic substitution and is generally<br />
expressed as a ratio of the rate constants,<br />
k light/k heavy. A normal isotope effect is one where<br />
the ratio of k light to k heavy is greater than 1. In an<br />
inverse isotope effect, the ratio is less than 1. A<br />
primary isotope effect is one which results from the<br />
making or breaking of a bond to an isotopically<br />
substituted atom and this must occur in the rate<br />
determining step. A secondary isotope effect is<br />
attributable to isotopic substitution of an atom not<br />
involved in bond making or breaking in the rate<br />
determining step. Thus when a hydrogen in a<br />
substrate is replaced by deuterium, there is often a<br />
change in the rate. Such changes are known as<br />
deuterium, isotope effects and are expressed by the<br />
ratio k H /k D , the typical value for this ratio is 7. The<br />
ground state vibrational energy (the zero-point<br />
vibrational energy) of a bond depends on the mass of<br />
the atoms and is lower when the reduced mass is<br />
higher. Consequently, D – C, D – O, D – N bonds,<br />
etc., have lower energies in the ground state than the<br />
corresponding H – C, H – O, H – N bonds, etc. Thus,<br />
complete dissociation of deuterium bond would<br />
require more energy than that for a corresponding<br />
hydrogen bond in the same environment. In case a H<br />
– C, H –O, or H – N bond is not broken at all in a<br />
reaction or is broken in a non-rate-determining step,<br />
substitution of deuterium for hydrogen generally does<br />
not lead to a change in the rate, however, if the bond<br />
is broken in the rate-determining step, the rate must<br />
be lowered by the substitution. This helps in<br />
determination of mechanism. In the bromination of<br />
acetone, the rate determining step is the<br />
tautomerization of acetone which involves cleavage<br />
of a C–H bond. In case this mechanistic assignment<br />
is correct, one should observe a substantial isotop<br />
effect on the bromination of deuterated acetone.<br />
Indeed k H /k D was found to be around 7.<br />
CH 3 COCH 3 + Br 2 ⎯→ CH 3 COCH 2 Br<br />
rate-determining step<br />
Bromoacetone<br />
OH<br />
CH 3 COCH 3 CH 3 C = CH 2<br />
Several mechanisms get support from kinetic isotope<br />
effect. Some of these are, oxidation of alcohols with<br />
chromic acid and electrophilic aromatic substitution.<br />
An example of a secondary isotope effect, where it is<br />
sure that the C – H bond does not break at all in the<br />
reaction. Secondary isotope effects for k H /k D are<br />
generally between 0.6 and 2.0.<br />
(CZ 3 ) 2 CHBr + H 2 O → (CZ 3 ) 2 CHOH + HBr<br />
the solvolysis of isopropyl bromide where Z = H or D, k H /k D is<br />
1.34 Secondary isotope effect.<br />
The substitution of tritium for hydrogen gives isotope<br />
effects which are numerically larger (k H /k T = 16).<br />
E2 elimination like S N 2 process takes place in one<br />
step (without the formation of any intermediates). As<br />
the attacking base begins to abstract a proton from a<br />
carbon next to the leaving group, the C – H bond<br />
begins to break, a new carbon-carbon double bond<br />
begins to form and leaving group begins to depart. In<br />
confirmation with this mechanism, the base induced<br />
elimination of HBr from (I) proceeds 7.11 times<br />
faster than the elimination of DBr from (II). Thus<br />
C–H or C – D bond is broken in the rate determining<br />
step. If it was not so there would not have been any<br />
rate difference.<br />
H<br />
Base<br />
– C – CH 2 Br – CH = CH 2<br />
H<br />
1-Bromo-2-phenylethane (I)<br />
D<br />
Faster reaction<br />
Base<br />
– C – CH 2 Br – CD = CH 2<br />
D<br />
Slower reaction<br />
1-Bromo-2,2-dideuterio-2-phenylethane (II)<br />
No deuterium isotope effect is found in E1 reactions<br />
since the rupture of C – H (or C – D) bond occurs<br />
after the rate determing step, rather than during it.<br />
Thus no rate difference can be measured between a<br />
deuterated and a non deuterated substrate.<br />
Mechanism Review : Substitution versus Elimination<br />
S N 2<br />
Primary substrate<br />
Back-side attack of Nu : with<br />
respect to LG<br />
Strong/polarizable unhindered<br />
nucleophile<br />
Bimolecular in ratedetermining<br />
step<br />
Concerted bond forming/bond<br />
breaking<br />
Inverse of stereochemistry<br />
Favored by polar aprotic<br />
solvent.<br />
S N 2 and E2<br />
Secondary or primary substrate<br />
Strong unhindered<br />
base/nucleophile leads to S N 2<br />
Strong hindered<br />
base/nucleophile leads to E2<br />
Low temperatrue (S N 2)/high<br />
temperature (E2)<br />
S N 1 and E 1<br />
Tertiary substrate<br />
Carbocation intermediate<br />
Weak nucleophile/base (e.g.,<br />
solvent)<br />
Unimolecular in ratedetermining<br />
step<br />
Racemization if S N 1<br />
Removal of β-hydrogen if E1<br />
Protic solvent assists ionization<br />
of LG<br />
Low temperature (S N 1)/high<br />
temperature (E2)<br />
E2<br />
Tertiary or secondary substrate<br />
Concerted anti-coplanar TS<br />
Bimolecular in ratedetermining<br />
step<br />
Strong hindered base<br />
High temperature<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 38 MAY 2010
UNDERSTANDING<br />
Physical Chemistry<br />
1. What is the solubility of AgCl in 0.20 M NH 3 ?<br />
Given : K sp (AgCl) = 1.7 × 10 –10 M 2<br />
K 1 = [Ag(NH 3 ) + ] / [Ag + ] [NH 3 ] = 2.33 × 10 3 M –1 and<br />
K 2 = [Ag(NH 3 ) + 2 ]/[Ag(NH 3 ) + ][NH 3 ] = 7.14 × 10 3 M –1<br />
Sol. If x be the concentration of AgCl in the solution, then<br />
[Cl – ] = x<br />
From the K sp for AgCl, we derive<br />
K −10 2<br />
[Ag + sp 1.7×<br />
10 M<br />
] = =<br />
−<br />
[Cl ] x<br />
If we assume that the majority of the dissolved Ag +<br />
goes into solution as Ag(NH 3 ) + 2 then [Ag(NH 3 ) + 2 ] = x<br />
Since two molecules of NH 3 are required for every<br />
Ag(NH 3 ) + 2 ion formed, we have [NH 3 ] = 0.20 M – 2x<br />
Therefore,<br />
⎛<br />
⎜<br />
.7× 10<br />
−<br />
+ 2<br />
[Ag ][NH ]<br />
x<br />
3<br />
K inst =<br />
=<br />
⎝<br />
+<br />
[Ag(NH3)<br />
2 ]<br />
= 6.0 × 10 –8 M 2<br />
From which we derive<br />
2<br />
2<br />
M ⎞<br />
⎟<br />
(0.20M − 2x)<br />
⎠<br />
x<br />
10<br />
1 2<br />
−8<br />
( 0.20M − 2x) 6.0×<br />
10 M<br />
=<br />
= 3.5 × 10 2<br />
2<br />
−10<br />
2<br />
x 1.7×<br />
10 M<br />
which gives x = [Ag(NH 3 ) + 2 ] = 9.6 × 10 –3 M, which<br />
is the solubility of AgCl in 0.20 M NH 3<br />
2. The values of Λ ∞ for HCl, NaCl and NaAc (sodium<br />
acetate) are 420, 126 and 91 Ω –1 cm 2 mol –1 ,<br />
respectively. The resistance of a conductivity cell is<br />
520 Ω when filled with 0.1 M acetic acid and drops<br />
to 122 Ω when enough NaCl is added to make the<br />
solution 0.1 M in NaCl as well. Calculate the cell<br />
constant and hydrogen-ion concentration of the<br />
solution. Given :<br />
∞<br />
Λ m (HCl) = 420 Ω –1 cm 2 mol –1 ,<br />
∞<br />
Λ m (NaCl) = 126 Ω –1 cm 2 mol –1 ,<br />
and Λ m (NaAc) = 91 Ω –1 cm 2 mol –1<br />
Sol. Resistance of 0.1 M HAc = 520 Ω<br />
Resistance of 0.1 M HAc + 0.1 M NaCl = 122 Ω<br />
Conductance due to 0.1 M NaCl,<br />
1 1<br />
G = – = 0.00627 Ω –1<br />
122 Ω 520 Ω<br />
Conductivity of 0.1 M NaCl solution<br />
k = Λ m c = (126 Ω –1 cm 2 mol –1 )(0.1 mol dm –3 )<br />
= 12.6 Ω –1 cm 2 dm –3 = 12.6 Ω –1 cm 2 (10 cm) –3<br />
= 0.0126 Ω –1 cm –1<br />
Cell constant,<br />
2<br />
−1<br />
−1<br />
k (0.0126 Ω cm )<br />
K = = = 2.01 cm –1<br />
G<br />
−1<br />
(0.00627Ω<br />
)<br />
Conductivity of 0.1 M HAc solution<br />
−1<br />
K 2.01 cm<br />
k = = R 520 Ω<br />
Molar conductivity of 0.1 M HAc solution<br />
−1<br />
−1<br />
k (2.01/ 520) Ω cm<br />
Λ m (HAc) = = c<br />
−3<br />
(0.1 mol dm )<br />
= 0.038 65 Ω –1 cm –1 dm 3 mol –1<br />
= 38.65 Ω –1 cm 2 mol –1<br />
According to Kohlrausch law, Λ ∞ (HAc) is given by<br />
∞<br />
Λ m (HAc) =<br />
∞<br />
Λ m (HCl) +<br />
∞<br />
Λ m (NaAc) –<br />
∞<br />
Λ m (NaCl)<br />
= (420 + 91 – 126) Ω –1 cm 2 mol –1<br />
= 385 Ω –1 cm 2 mol –1<br />
Therefore, the degree of dissociation of acetic acid is<br />
given as<br />
α =<br />
Λ<br />
Λ<br />
m =<br />
∞<br />
m<br />
(38.65 Ω<br />
(385Ω<br />
−1<br />
−1<br />
cm<br />
cm<br />
2<br />
2<br />
mol<br />
mol<br />
−1<br />
−1<br />
)<br />
)<br />
≈ 0.1<br />
and the hydrogen-ion concentration of 0.1 M HAc<br />
solution is<br />
[H + ] = cα = (0.1 M)(0.1) = 0.01 M<br />
Thus, its pH is pH = – log{[H + ]/M} = – log(0.01) = 2<br />
3. Potassium alum is KA1(SO 4 ) 2 .12H 2 O. As a strong<br />
electrolyte, it is considered to be 100% dissociated<br />
into K + , Al 3+ , and SO 2– 4 . The solution is acidic<br />
because of the hydrolysis of Al 3+ , but not so acidic as<br />
might be expected, because the SO 2– 4 can sponge up<br />
some of the H 3 O + by forming HSO – 4 . Given a<br />
solution made by dissolving 11.4 g of<br />
KA1(SO 4 ) 2 .12H 2 O in enough water to make 0.10 dm 3<br />
of solution, calculate its [H 3 O + ] :<br />
(a) Considering the hydrolysis<br />
Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O +<br />
with K h = 1.4 × 10 –5 M<br />
(b) Allowing also for the equilibrium<br />
HSO – 4 + H 2 O H 3 O + 2–<br />
+ SO 4<br />
with K 2 = 1.26 × 10 –2 M<br />
11.4 g<br />
Sol. (a) Amount of alum =<br />
= 0.024 mol<br />
−1<br />
474.38 g mol<br />
0.024 mol<br />
Molarity of the prepared solution =<br />
3<br />
0.1dm<br />
= 0.24 M<br />
Hydrolysis of Al 3+ is<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 39 MAY 2010
Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O +<br />
2+<br />
+<br />
[Al(OH) ][H3O<br />
]<br />
K h =<br />
3+<br />
[Al ]<br />
If x is the concentration of Al 3+ that has hydrolyzed,<br />
we have<br />
(x)(x)<br />
K h =<br />
0.24M − x<br />
= 1.4 × 10 –5 M<br />
Solving for x, we get<br />
[H 3 O + ] = x = 1.82 × 10 –3 M<br />
(b) We will have to consider the following equilibria.<br />
Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O +<br />
H 3 O + 2–<br />
+ SO 4 HSO – 4 + H 2 O<br />
Let z be the concentration of SO 2– 4 that combines<br />
with H 3 O + and y be the net concentration of H 3 O +<br />
that is present in the solution. Since the concentration<br />
z of SO 2–<br />
4 combines with the concentration z of<br />
H 3 O + , it is obvious that the net concentration of H 3 O +<br />
produced in the hydrolysis reaction of Al 3+ is (y + z).<br />
Thus, the concentration (y + z) of Al 3+ out of 0.24 M<br />
hydrolyzes in the solution. With these, the<br />
concentrations of various species in the solution are<br />
3+<br />
2<br />
Al + 2H 2 O Al(OH)<br />
+ 3 O 0.24 M−y−z<br />
y+<br />
z<br />
y<br />
H 3 O 2−<br />
+ SO 4 HSO −<br />
4 + H 2 O<br />
y 0.48 M−z<br />
z<br />
(y + z)(y)<br />
Thus, K h =<br />
= 1.4 × 10 –5 M ...(i)<br />
(0.24M − y − z)<br />
z<br />
1<br />
K 2 =<br />
=<br />
...(ii)<br />
−2<br />
y(0.48M − z) 1.26×<br />
10 M<br />
From Eq. (ii), we get<br />
(0.48M)y<br />
z =<br />
2<br />
(1.26× 10<br />
− M) + y<br />
Substituting this in Eq. (i), we get<br />
⎛<br />
⎞<br />
⎜<br />
(0.48M)y<br />
⎟<br />
y +<br />
y<br />
−2<br />
⎝ (1.26×<br />
10 M) + y ⎠<br />
= 1.4 × 10 –5<br />
⎛<br />
⎞<br />
⎜<br />
(0.48M)y<br />
⎟<br />
0.24 − y −<br />
−2<br />
⎝ (1.26×<br />
10 M) + y ⎠<br />
Making an assumption that y
`tà{xÅtà|vtÄ V{tÄÄxÇzxá<br />
1 Set<br />
This section is designed to give <strong>IIT</strong> <strong>JEE</strong> aspirants a thorough grinding & exposure to variety<br />
of possible twists and turns of problems in mathematics that would be very helpful in facing<br />
<strong>IIT</strong> <strong>JEE</strong>. Each and every problem is well thought of in order to strengthen the concepts and<br />
we hope that this section would prove a rich resource for practicing challenging problems and<br />
enhancing the preparation level of <strong>IIT</strong> <strong>JEE</strong> aspirants.<br />
By : Shailendra Maheshwari<br />
Solutions will be published in next issue<br />
Joint Director Academics, <strong>Career</strong> <strong>Point</strong>, Kota<br />
Passage :<br />
A bag contains ‘n’ cards marked 1, 2, 3, ......, n. ‘X’<br />
draws a card from the bag and the card is put back<br />
into the bag. Then ‘Y’ draws a card. The probability<br />
that ‘X’ draws.<br />
1. The same card as ‘Y’ is –<br />
1<br />
1<br />
(A) (B) n 2n<br />
(C)<br />
1<br />
2<br />
n<br />
(D) n<br />
2<br />
2. a higher card than ‘Y’ is –<br />
(A)<br />
(C)<br />
n −1<br />
n<br />
n −1<br />
2<br />
n<br />
(B)<br />
(D)<br />
3. a lower card than ‘Y’ is –<br />
n −1<br />
(A)<br />
(B)<br />
n<br />
(C)<br />
n −1<br />
2<br />
n<br />
1<br />
0<br />
1<br />
0<br />
∫ x<br />
4. Evaluate : 228<br />
∫ x<br />
25<br />
24<br />
(d)<br />
(1 − x)<br />
(1 − x)<br />
5. Find the minimum value of<br />
⎛<br />
2<br />
(x 1 – x 2 ) 2 + ⎜<br />
x<br />
− (17 − x<br />
⎝ 20<br />
n −1<br />
2n<br />
n −1<br />
2<br />
2n<br />
n −1<br />
2n<br />
n − 1<br />
2<br />
2n<br />
50<br />
49<br />
dx<br />
dx<br />
= ?<br />
1<br />
2)(x<br />
2 −13<br />
)<br />
where x 1 ∈ R + and x 2 ∈ (13, 17).<br />
⎞<br />
⎟<br />
⎠<br />
2<br />
6. Let f(x) = a 1 tan x + a 2 tan 2<br />
x + a3 tan 3<br />
x + ...... + an<br />
tan n<br />
x , where a1 , a 2 , a 3 , ... a n ∈ R and n ∈ N. If | f(x) |<br />
n<br />
≤ | tan x | for ∀ x ∈ ⎛ π π ⎞<br />
a<br />
⎜−<br />
, ⎟ , Prove that ∑<br />
⎝ 2 2 ⎠ i=<br />
1<br />
i<br />
i<br />
≤ 1<br />
7. Let az 2 + bz + c be a polynomial with complex<br />
coefficients such that a and b are non zero. Prove that<br />
the zeros of this polynomial lie in the region.<br />
b c<br />
| z | ≤ + + a b<br />
8. Find the fifth degree polynomial which leaves<br />
remainder 1 when divided by (x – 1) 3 and remainder<br />
–1 when divided by (x + 1) 3 .<br />
9. A quadrilateral ABCD is inscribed in a circle of<br />
radius R such that AB 2 + CD 2 = 4R 2 . Using vector<br />
method prove that its diagonals are at right angle.<br />
10. Through a focus of an ellipse two chords are drawn<br />
and a conic is described to pass through their<br />
extremities, and also through the centre of the ellipse.<br />
Prove that it cuts the major axis in another fixed<br />
point.<br />
High Speed Avalanches<br />
Although an avalanche can mean the<br />
fall of any material e.g. snow, soil and<br />
even rocks, in common usage it<br />
generally refers to a falling mass of<br />
ice and snow which breaks away<br />
from the side of a mountain or cliff<br />
and surges down at great speed.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 41 MAY 2010
θ 2<br />
θ1<br />
MATHS<br />
Students' Forum<br />
Expert’s Solution for Question asked by <strong>IIT</strong>-<strong>JEE</strong> Aspirants<br />
1. An ellipse of eccentricity 2/3 is inscribed in an ellipse<br />
of equal eccentricity and area equals to 9 square units<br />
in such a way that both the ellipses touch each other<br />
at one end of their common major axis. If length of<br />
major axis of smaller ellipse is equal to length of<br />
minor axis of bigger ellipse, find the area of the<br />
bigger ellipse outside the smaller ellipse.<br />
Sol. The required figure will be drawn as follows<br />
y<br />
2. Given a point P on the circumference of the circle<br />
|z| = 1, and vertices A 1 , A 2 , ......, A n of an inscribed<br />
regular polygon of n sides. Prove using complex<br />
numbers that<br />
(PA 1 ) 2 + (PA 2 ) 2 + ......... + (PA n ) 2 is a constant.<br />
Sol. Without loss of generality we can take P as<br />
1 + 0i.<br />
i.e., P ≡ C is 0<br />
A 3<br />
A 2<br />
A 1<br />
x<br />
θ n<br />
P<br />
A n<br />
and we can redraw the figure for our purpose (i.e.<br />
keeping the area out side the smaller ellipse and<br />
inside the bigger ellipse same) as<br />
(b, 0)<br />
(c, 0)<br />
(–a, 0) (–b, 0) (–c, 0)<br />
(–b, 0)<br />
Therefore, we can let the ellipses be<br />
2<br />
2<br />
2<br />
y<br />
(b, 0) (a, 0)<br />
x<br />
2 2<br />
x<br />
2 +<br />
2<br />
x y<br />
and +<br />
2 = 1<br />
b c<br />
Required area = π ab – π bc<br />
= π b (a – c)<br />
Now b 2 = a 2 (1 – e 2 ) and c 2 = b 2 (1 – e 2 ) (1 – e 2 ) 2<br />
⇒ c = a (1 – e 2 )<br />
⇒ a – c = ae 2<br />
Thus required area = πb (ae 2 )<br />
= πabe 2<br />
2<br />
⎛ 2 ⎞<br />
= 9 × ⎜ ⎟⎠ = 4 sq. units.<br />
⎝ 3<br />
a<br />
y = 1<br />
b<br />
Let A r ≡ C is θ r , r = 1, 2, ......, n.<br />
PA r = |Cis θ r – Cis 0| = |(cosθ r – 1) + i(sinθ r )|<br />
PA 2 r = (cos θ r – 1) 2 + (sinθ r ) 2<br />
= 2 – 2cos θ r<br />
n<br />
⇒ ∑<br />
2<br />
( PA r ) = 2n – 2<br />
r=<br />
1<br />
∑ cos θr<br />
r=<br />
1<br />
n<br />
⎡<br />
n<br />
⎤<br />
Now, ∑ cos θr<br />
= Re ⎢∑Cis<br />
θr<br />
⎥<br />
r=<br />
1<br />
⎢⎣<br />
r=<br />
1 ⎥⎦<br />
iθ1 iθ2<br />
iθn<br />
= Re [ e + e + ....... + e ]<br />
⎡ ⎛<br />
n<br />
⎞⎤<br />
⎢ ⎜ ⎛<br />
2π<br />
i ⎞<br />
iθ<br />
⎟⎥<br />
⎢<br />
e ⎜ ⎟<br />
1<br />
⎜1−<br />
e<br />
n<br />
⎟⎥<br />
⎢<br />
⎜ ⎟<br />
⎥<br />
= Re<br />
⎝ ⎝ ⎠ ⎠<br />
⎢<br />
2π<br />
⎥<br />
⎢<br />
i ⎥<br />
⎢ 1−<br />
e<br />
n<br />
⎥<br />
⎢<br />
⎥<br />
⎣<br />
⎦<br />
Q θ 2 – θ 1 = θ 3 – θ 2 = ..... = θ n – θ n–1 =<br />
⎡<br />
iθ<br />
⎤<br />
= Re<br />
⎢e 1 (1 −1)<br />
⎥<br />
= 0<br />
⎢ 2π<br />
⎥<br />
i<br />
⎢⎣<br />
1−<br />
e<br />
n ⎥⎦<br />
n<br />
Hence, ∑<br />
=<br />
r 1<br />
2<br />
( PA r ) = 2n = constant.<br />
n<br />
2π<br />
n<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 42 MAY 2010
3. In a class of 20 students, the probability that exactly x<br />
students pass the examination is directly proportional<br />
to x 2 (0 ≤ x ≤ 20). Find out the probability that a<br />
student selected at random has passed the<br />
examination. If a selected students has been found to<br />
pass the examination find out the probability that<br />
he/she is only student to have passed the<br />
examination.<br />
Sol. Let E x : event that exactly x out of 20 students<br />
pass the examination<br />
and A : event that a particular student passes<br />
the examination<br />
⇒ P(E x ) = kx 2 (k is the proportionality constant)<br />
Now, E 0 , E 2 , ....., E 20 are mutually exclusive and<br />
exhaustive events.<br />
⇒ P(E 0 ) + P(E 1 ) + P(E 2 ) + ... + P(E 20 ) = 1<br />
⇒ 0 + k(1) 2 + k(2) 2 + .... + k(20) 2 = 1<br />
⎡ (20)(20 + 1)(40 + 1) ⎤<br />
⇒ k ⎢<br />
⎥ = 1<br />
⎣ 6 ⎦<br />
1<br />
⇒ k = 2870<br />
20<br />
Now, P(A) = ∑<br />
=<br />
20<br />
= ∑<br />
=<br />
x<br />
0<br />
x<br />
0<br />
P (E<br />
2 x<br />
kx . = 20<br />
x ).P(A / E x )<br />
k<br />
20<br />
20<br />
∑<br />
x=<br />
0<br />
1 ⎡ 20(20 + 1) ⎤ 63<br />
=<br />
20 2870<br />
⎢<br />
2<br />
⎥ =<br />
× ⎣ ⎦ 82<br />
P(E1<br />
).P(A / E1)<br />
and P(E 1 /A) =<br />
P(A)<br />
1 1<br />
(1)<br />
2 .<br />
=<br />
2870 20 1<br />
=<br />
63 44100<br />
82<br />
4. Find the set of values of ‘a’ for which minimum<br />
value of x 3 – 6ax 2 + 9a 2 x + 7, x ∈ [–1, 2] is 3.<br />
Sol. Let f(x) = x 3 – 6ax 2 + 9a 2 x + 7<br />
a ≠ 0, otherwise f(x) = x 3 + 7, which is always<br />
increasing and hence min f = f(–1) = 6 ≠ 3.<br />
Now f´(x) = 3x 2 – 12ax + 9a 2 = 0 for stationary points<br />
⇒ x = a, 3a<br />
CaseI : a > 0<br />
⇒ –1 is always in the left of a.<br />
Case I. (a) : 2 ≤ a, then<br />
3 = min f = f(–1) = –1 – 6a – 9a 2 + 7<br />
⇒ 3a 2 + 2a – 1 = 0, no admissible value of a is<br />
obtained.<br />
(a,f(a))<br />
x<br />
2<br />
3<br />
(3a,f(3a))<br />
Case I. (b) : –1 < a < 2 < 3a<br />
i.e., 3<br />
2 < a < 2, then<br />
3 = min f = min{f(–1), f(2)}<br />
= min {–1 –6a – 9a 2 + 7, 8 – 24 a + 18a 2 + 7}<br />
= –1 – 6a – 9a 2 + 7<br />
as – 1 – 6a – 9a 2 + 7 < 8 – 24a + 18a 2 + 7<br />
i.e., 3a 2 – 2a + 1 > 0, which is true<br />
Hence 3 = –1 – 6a – 9a 2 + 7<br />
⇒ a = –1 or 3<br />
1 , none of which is possible.<br />
Case I(c) : 3a ≤ 2<br />
⇒ 3 = min f = min{f(–1), f(3a)}<br />
= {–1 – 6a –9a 2 + 7, 18a 3 + 7}<br />
= –1 – 6a – 9a 2 + 7,<br />
as 18a 3 + 77 – 1 – 6a – 9a 2 + 7<br />
i.e., 18a 3 + 9a 2 + 6a + 170<br />
which is true as a > 0. Hence a = – 1 or 3<br />
1 ,<br />
in which a = 3<br />
1 is permissible.<br />
Case II : a < 0<br />
⇒ 2 is always in the right of a<br />
Case II (a) a ≤ –1<br />
⇒ 3 = min f = f(–1)<br />
⇒ a = –1, as a = 3<br />
1<br />
Hence a = –1 is one possibility<br />
(3a,f(3a))<br />
(a,f(a))<br />
Case II (b) 3a ≤ –1 < a<br />
1<br />
⇒ –1, as a = 3<br />
⇒ 3 = min f = f(a) = a 3 – 6a 3 + 9a 3 + 7<br />
⇒ 4a 3 = – 4 ⇒ a = – 1, not possible<br />
Case II(c) –1 < 3a ⇒ 3 = min f = min {f(–1), f(a) }<br />
= min {–1 –6a – 9a 2 + 7, 4a 3 + 7}<br />
= 4a 3 + 7,<br />
as 4a 3 + 7 < –1 – 6a – 9a 2 + 7<br />
as (a + 1) 2 (4a + 1) < 0. Hence a = –1, not possible<br />
Hence a = –1 or a = 1/3<br />
5. A man standing at a distance 5m in front of the base<br />
of a building 10m high on which a flagstaff is<br />
mounted observes that the top of the building and the<br />
top of a mountain behind the building are along the<br />
same straight line. When he recedes by a distance of<br />
48 m he observes that now the top of the flagstaff and<br />
the top of the mountain are along the same straight<br />
line. If at both the locations, the flagstaff subtends the<br />
same angle at the man’s eye, find the height of<br />
mountain.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 43 MAY 2010
Sol. CD : Flagstaf<br />
DE : Building<br />
KF : Mountain (height = h say)<br />
The figure illustrates the situation.<br />
Since, ∠CBD = ∠CAD = α say, points A, B, C and<br />
D are concyclic.<br />
⇒ ∠ABD = ∠ACD = 90º – (α + β)<br />
⇒ ∠ABC = 90º – (α + β) + α = 90º – β = ∠KCH<br />
K<br />
90º – β<br />
α 10<br />
α<br />
90º – β<br />
β<br />
B 48<br />
Now, h = KH + HF<br />
A 5 E<br />
F<br />
= (CH) tan (90º – β) + (BE) tan(90º – β)<br />
(Q HF = CE)<br />
= [DG + (BA + AE) cot β<br />
= [KG cot β + (48 + 5)] cot β<br />
⇒ h = [(h – 10)cotβ + 53] cot β<br />
(Q KG = KF – GF)<br />
5<br />
Putting cot β = 10<br />
1 = , we get 2<br />
⎛ h −10<br />
⎞ 1<br />
h = ⎜ + 53⎟ ⎝ 2 ⎠ 2<br />
⇒ 4h = h – 10 + 106<br />
⇒ 3h = 96<br />
⇒ h = 32 m<br />
6. If a, b, c and n are positive integers such that<br />
a + b + c = n, show that<br />
(a a b b c c ) 1/n + (a b b c c a ) 1/n + (a c b a c b ) 1/n ≤ n.<br />
Sol. Since a, b, c are integers, from A.M. – G.M.<br />
inequality we can write<br />
(a + a + ....a times) + (b + b + ...b times) + (c + c + ...c times)<br />
a + b + c<br />
1/a +b+c<br />
≥ [(a.a....a times)(b.b....b times)(c.c...c times)]<br />
1<br />
a.a + b.b + c.c a b c<br />
⇒<br />
≥ ( a b c ) a+<br />
b+<br />
c<br />
a + b + c<br />
1<br />
c.a + a.b + b.c c a c<br />
Similarly,<br />
≥ ( a b c ) a+<br />
b+<br />
c<br />
c + a + b<br />
1<br />
b.a + c.b + a.c b c a<br />
and<br />
≥ ( a b c ) b+<br />
c+<br />
a<br />
b + c + a<br />
Adding these three inequalities, we get<br />
2 2 2<br />
a + b + c + 2ab + 2bc + 2ca<br />
≥ (a a b b c c ) 1/n<br />
a + b + c<br />
+ (a c b a c b ) 1/n + (a b b c c a ) 1/n<br />
2<br />
(a + b + c)<br />
where LHS =<br />
= a + b + c Hence proved<br />
a + b + c<br />
C<br />
D<br />
β<br />
H<br />
G<br />
How do Satellites<br />
Stay Up?<br />
Satellites orbit the earth because of the force of<br />
gravity. To understand why this happens and why the<br />
satellite does not get pulled in and fall, we have to<br />
understand what forces do. A force will change the motion<br />
of an object; it might speed it up, slow it down or change<br />
its direction. For example, if you are running and someone<br />
pushes you from behind, you speed up (the force is in the<br />
direction of your motion). But if someone pushes you in<br />
the chest when you are running, you slow down (the force<br />
is in the opposite direction to your motion). If you are<br />
running and someone pushes you from the side, you move<br />
away from them, changing your direction. (the force is at<br />
right angles to the motion). This idea is called Newton’s<br />
First Law.<br />
To make something move in a circle it must be<br />
moving and have a force that is always at right angles to the<br />
motion so that it constantly changes direction. This force is<br />
called the centripetal force.<br />
Imagine swinging a rock on a string around your<br />
head. The tension in the string pulls the rock round in a<br />
circle (this is the force at right angles to the motion). So<br />
the tension is the centripetal force. If we cut the string, the<br />
rock will continue in a straight line because there is no<br />
longer a force to change its direction. For a satellite, the<br />
centripetal force is the gravitational force, the pull of the<br />
earth. If we could switch gravity off, we would lose all our<br />
satellites as they move off in straight lines!<br />
Going back to the rock example, we need to put<br />
energy into keeping the rock moving because the rock is<br />
moving through air and is losing energy constantly because<br />
of air resistance. We don’t need to do this with satellites<br />
because they are moving through space where there is no<br />
air, so no air resistance acts on the satellites and they don’t<br />
slow down.<br />
Many people think there is a centrifugal force<br />
acting which pulls the satellite (or rock) outwards. This is<br />
not the case; there is no such thing as a centrifugal force.<br />
Imagine riding in the back seat of a car as it turns the<br />
corner. Let us assume the back seat is very slippery and<br />
you don’t have a seatbelt on. As the car turns the corner,<br />
you slide from the inside to the outside. You could argue<br />
that a force is pushing you outwards. In fact the reason you<br />
move outwards is that there is no force keeping you<br />
moving in the same circle as the car. What you really do is<br />
continue in a straight line. Eventually you will hit the far<br />
door and the door then pushes you providing the<br />
centripetal force to keep you moving in the same circle as<br />
the car. If the car door was open you would not have a<br />
centripetal force acting and you would continue in a<br />
straight line out of the car! (Don’t try this at home!)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 44 MAY 2010
MATHS<br />
COMPLEX NUMBER<br />
Mathematics Fundamentals<br />
− 1 is denoted by ‘i’ and is pronounced as ‘iota’.<br />
i = − 1 ⇒ i 2 = –1, i 3 = –i, i 4 = 1.<br />
If a, b ∈ R and i = − 1 then a + ib is called a<br />
complex number. The complex number a + ib is also<br />
denoted by the ordered pair (a, b)<br />
If z = a + ib is a complex number, then :<br />
(i) a is called the real part of z and we write<br />
Re (z) = a.<br />
(ii) b is called the imaginary part of z and we write<br />
Im (z) = b<br />
Two complex numbers z 1 and z 2 are said to be equal<br />
complex numbers if Re (z 1 ) = Re (z 2 ) and Im (z 1 ) =<br />
Im (z 2 ).<br />
If z = x + iy is a non zero complex number, then 1/z<br />
is called the multiplicative inverse of z.<br />
If x + iy is a complex number, then the complex<br />
number x – iy is called the conjugate of the complex<br />
number x + iy and we write x + iy = x – iy.<br />
Algebra of Complex Numbers<br />
(i) Addition : (a + ib) + (c + id) = (a + c) + i(b + d)<br />
(ii) Subtraction :<br />
(a + ib) – (c + id) = (a – c) + i(b – d)<br />
(iii) Multiplication :<br />
(a + ib) + (c + id) = (ac – bd) + i(ab + bc)<br />
(iv) Division by a non-zero complex number :<br />
a + ib ac + bd bc − ad<br />
= + i , (c + id) ≠ 0<br />
2 2 2 2<br />
c + id c + d c + d<br />
Properties : If z 1 , z 2 are complex numbers, then<br />
(i) ( z 1 ) = z 1<br />
(ii) z + z = 2 Re (z)<br />
(iii) z – z = 2i Im (z)<br />
(iv) z = z iff z is purely real<br />
(v) z = z iff z is purely imaginary<br />
(vi) z 1 + z2<br />
= z1<br />
+ z2<br />
(vii) z 1 – z2<br />
= z1<br />
– z2<br />
(viii) z 1 .z2<br />
= z1<br />
. z2<br />
⎛ z<br />
(ix)<br />
⎜<br />
⎝ z<br />
1<br />
2<br />
⎞<br />
⎟ =<br />
⎠<br />
z1<br />
z2<br />
provided z 2 ≠ 0<br />
If x + iy is a complex number, then the non-negative<br />
2 2<br />
ral number x + y is called the modulus of the<br />
complex number x + iy and write<br />
| x + iy| =<br />
2 2<br />
x + y<br />
Properties : If z 1 , z 2 are complex numbers, then<br />
(i) | z 1 | = 0 iff z 1 = 0<br />
(ii) | z 1 | = | z 1 | = | – z 1 |<br />
(iii) – | z 1 | ≤ Re (z 1 ) ≤ | z 1 |<br />
(iv) – | z 1 | ≤ Im (z 1 ) ≤ | z 1 |<br />
(v) | z 1 z 1 | = | z 1 | 2<br />
(vi) | z 1 + z 2 | ≤ | z 1 | + | z 2 |<br />
(vii) | z 1 – z 2 | ≥ | z 1 | – | z 2 |<br />
(viii) | z 1 z 2 | = | z 1 | | z 2 |<br />
z1<br />
| z<br />
(ix) =<br />
1 |<br />
, provided z 2 ≠ 0<br />
z2<br />
| z2<br />
|<br />
(x) | z 1 + z 2 | 2 = | z 1 | 2 + | z 2 | 2 + 2 Re (z 1 z 2 )<br />
(xi) | z 1 – z 2 | 2 = | z 1 | 2 + | z 2 | 2 – 2 Re (z 1 z 2 )<br />
(xi) | z 1 + z 2 | 2 + | z 1 – z 2 | 2 = 2 [| z 1 | 2 + | z 2 | 2 ].<br />
De Moivre’s Theorem<br />
(i) If n is any integer (positive or negative), then<br />
(cos θ + i sin θ) n = cos nθ + i sin nθ<br />
(ii) If n is a rational number, then the value or one of<br />
the values of (cos θ + i sin θ) n is cos nθ + i sin nθ<br />
Euler’s Formula<br />
e iθ = cos θ + i sin θ and e –iθ = cos θ – i sin θ<br />
Square root of complex number<br />
Square root of z = a + ib are given by<br />
⎡<br />
± ⎢<br />
⎢⎣<br />
⎡<br />
± ⎢<br />
⎢⎣<br />
⎛ | z | + a ⎞<br />
⎜ ⎟ + i<br />
⎝ 2 ⎠<br />
⎛ | z | +<br />
⎜<br />
⎝ 2<br />
a<br />
⎞<br />
⎟<br />
⎠<br />
⎛ | z | −a<br />
⎞⎤<br />
⎜ ⎟⎥<br />
for b > 0 and<br />
⎝ 2 ⎠⎥⎦<br />
⎛ | z | −a<br />
⎞⎤<br />
– i ⎜ ⎟⎥<br />
for b < 0.<br />
⎝ 2 ⎠⎥⎦<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 45 MAY 2010
−1+ i 3<br />
If ω = , then the cube roots of unity are 1, ω<br />
2<br />
and ω 2 . We have:<br />
(i) 1 + ω + ω 2 = 0 (ii) ω 3 = 1<br />
Let z = x + iy be any complex number.<br />
Let z = r (cos θ + i sin θ) where r > 0.<br />
∴ x = r cos θ and y = r sin θ<br />
∴ x 2 + y 2 = r 2<br />
⇒ r =<br />
∴ cos θ =<br />
x<br />
2<br />
2<br />
y<br />
x + (Q r > 0)<br />
2<br />
x<br />
+ y<br />
2<br />
and sin θ =<br />
x<br />
2<br />
y<br />
+ y<br />
The value of θ is found by solving these equations. θ<br />
is called the argument (or amplitude) of z.<br />
If – p < θ ≤ π, then θ is called the principal argument<br />
of z.<br />
Identification of θ –<br />
x y arg(z) Interval of θ<br />
⎛ π ⎞<br />
+ + θ ⎜0<br />
< θ < ⎟<br />
⎝ 2 ⎠<br />
⎛ – π ⎞<br />
+ – –θ ⎜ < θ < 0⎟ ⎝ 2 ⎠<br />
⎛ π ⎞<br />
– + (π – θ) ⎜ < θ < π⎟ ⎝ 2 ⎠<br />
⎛ – π ⎞<br />
– – –(π – θ) ⎜ – π < θ < ⎟<br />
⎝ 2 ⎠<br />
If z 1 and z 2 are two complex numbers then<br />
(i) | z 1 – z 2 | is the distance between the points with<br />
affixes z 1 and z 2 .<br />
mz<br />
(ii)<br />
2 + nz1<br />
is the affix of the point dividing the<br />
m + n<br />
line joining the points with affixes z 1 and z 2 in the<br />
ratio m : n internally.<br />
mz<br />
(iii)<br />
2 – nz1<br />
is the affix of the point dividing the<br />
m – n<br />
line joining the points with affixes z 1 and z 2 in the<br />
ratio m : n externally where m ≠ n.<br />
(iv) If z 1 , z 2 , z 3 are the affixes of the vertices of a<br />
z1 + z2<br />
+ z3<br />
triangle then the affix of its centroid is<br />
.<br />
3<br />
(v) z = tz 1 + (1 – t)z 2 is the equation of the line<br />
joining points with affixes z 1 and z 2 . Here ‘t’ is a<br />
parameter.<br />
2<br />
z − z<br />
(vi)<br />
1 z − z<br />
=<br />
1<br />
is the equation of the line<br />
z2<br />
− z1<br />
z2<br />
− z1<br />
joining points with affixes z 1 and z 2 .<br />
Three points with affixes z 1 , z 2 , z 3 are collinear if<br />
z1<br />
z1<br />
1<br />
z2<br />
z2<br />
1 = 0.<br />
z3<br />
z3<br />
1<br />
The general equation of a straight line is<br />
a z + az + b = 0 , where b is any real number.<br />
(i) | z – z 1 | < r represents the circle with centre z 1<br />
and radius r.<br />
(ii) | z – z 1 | < r represents the interior of the circle<br />
with centre z 1 and radius r.<br />
z − z1<br />
= k represents a circle line which is the<br />
z − z1<br />
perpendicular bisector of the line segment joining<br />
points with affixes z 1 and z 2 .<br />
(z – z 1 ) ( z − z2)<br />
+ ( z − z1)<br />
+ (z – z 2 ) = 0 represents<br />
the circle with line joining points with affixes z 1 and<br />
z 2 as a diameter.<br />
| z – z 1 | + | z – z 2 | = 2k, k ∈ R + represents the ellipse<br />
with foci at points with affixes z 1 and z 2 .<br />
If z 1 , z 2 , z 3 be the affixes of the points A, B, C<br />
respectively, then the angle between AB and AC is<br />
⎛ z ⎞<br />
given by arg<br />
⎜<br />
3 − z1<br />
⎟ .<br />
⎝ z2<br />
− z1<br />
⎠<br />
If z 1 , z 2 , z 3 , z 4 are the affixes of the points A, B, C, D<br />
respectively, then the angle between AB and CD is<br />
⎛ z ⎞<br />
given by arg ⎜ 2 − z1<br />
⎟<br />
.<br />
⎝ z4<br />
− z3<br />
⎠<br />
nth roots of a complex number<br />
Let z = r (cos θ + i sin θ), r > 0 be any complex<br />
number. nth root o z = z 1/n<br />
= r 1/n ⎛ 2kπ + θ 2kπ + θ ⎞<br />
⎜cos + isin ⎟ ,<br />
⎝ n<br />
n ⎠<br />
where k = 0, 1, 2, ………, n – 1.<br />
There are n distinct values and sum of all these<br />
values is 0.<br />
Logarithm of a complex number<br />
Let z = re iθ be any complex number.<br />
Then log z = log re iθ = log r + log e iθ<br />
= log r + iθ log e = log r + iθ.<br />
∴ log z = log | z | + i amp (z).<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 46 MAY 2010
MATHS<br />
MATRICES &<br />
DETERMINANTS<br />
Mathematics Fundamentals<br />
Matrices :<br />
An m × n matrix is a rectangular array of mn<br />
numbers (real or complex) arranged in an ordered set<br />
of m horizontal lines called rows and n vertical lines<br />
called columns enclosed in parentheses. An m × n<br />
matrix A is usually written as :<br />
⎡ a<br />
⎢<br />
⎢<br />
a<br />
⎢ M<br />
A = ⎢<br />
⎢ a<br />
⎢ M<br />
⎢<br />
⎢⎣<br />
a<br />
11<br />
21<br />
i1<br />
m1<br />
a12<br />
a 22<br />
M<br />
ai2<br />
M<br />
a m2<br />
...<br />
...<br />
...<br />
...<br />
a1j<br />
a 2 j<br />
aij<br />
a mj<br />
...<br />
...<br />
...<br />
...<br />
a1n<br />
a 2n<br />
ain<br />
a mn<br />
⎤<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥<br />
⎥⎦<br />
Where 1 ≤ i ≤ m and 1 ≤ j ≤ n<br />
and is written in compact form as A = [a ij ] m× n<br />
A matrix A = [a ij ] m × n is called<br />
(i) a rectangular matrix if m ≠ n<br />
(ii) a square matrix if m = n<br />
(iii) a row matrix or row vector if m = 1<br />
(iv) a column matrix or column vector if n = 1<br />
(v) a null matrix if a ij = 0 for all i, j and is denoted by<br />
O m× n<br />
(vi) a diagonal matrix if a ij = 0 for i ≠ j<br />
(vii) a scalar matrix if a ij = 0 for i ≠ j and all diagonal<br />
elements a ii are equal<br />
Two matrices can be added only when thye are of same<br />
order. If A = [a ij ] m × n and B = [b ij ] m × n , then sum of A<br />
and B is denoted by A + B and is a matrix [a ij + b ij ] m × n<br />
The product of two matrices A and B, written as AB,<br />
is defined in this very order of matrices if number of<br />
columns of A (pre factor) is equal to the number of<br />
rows of B (post factor). If AB is defined , we say that<br />
A and B are conformable for multiplication in the<br />
order AB.<br />
If A = [a ij ] m × n and B = [b ij ] n × p , then their product AB<br />
is a matrix C = [c ij ] m × p where<br />
C ij = sum of the products of elements of ith row of A<br />
with the corresponding elements of jth column of B.<br />
Types of matrices :<br />
(i) Idempotent if A 2 = A<br />
(ii) Periodic if A k+1 = A for some positive integer k.<br />
The least value of k is called the period of A.<br />
(iii) Nilpotent if A k = O when k is a positive integer.<br />
Least value of k is called the index of the<br />
nilpotent matrix.<br />
(iv) Involutary if A 2 = I.<br />
The matrix obtained from a matrix A = [a ij ] m × n by<br />
changing its rows into columns and columns of A into<br />
rows is called the transpose of A and is denoted by A′.<br />
A square matrix a = [a ij ] n × n is said to be<br />
(i) Symmetric if a ij = a ji for all i and j i.e. if A′ = A.<br />
(ii) Skew-symmetric if<br />
a ij = – a ji for all i and j i.e., if A′ = –A.<br />
Every square matrix A can be uniquely written as sum<br />
of a symmetric and a skew-symmetric matrix.<br />
A = 2<br />
1 (A + A′) + 2<br />
1 (A – A′) where 2<br />
1 (A + A′) is<br />
symmetric and 2<br />
1 (A – A′) is skew-symmetric.<br />
Let A = [a ij ] m × n be a given matrix. Then the matrix<br />
obtained from A by replacing all the elements by their<br />
conjugate complex is called the conjugate of the matrix<br />
A and is denoted by A = [a ] .<br />
Properties :<br />
(i) ( A ) = A<br />
(ii) ( A + B)<br />
= A + B<br />
(iii) (λ A)<br />
= λ A , where λ is a scalar<br />
(iv) ( A B) = A B .<br />
Determinant :<br />
Consider the set of linear equations a 1 x + b 1 y = 0 and<br />
a 2 x + b 2 y = 0, where on eliminating x and y we get<br />
the eliminant a 1 b 2 – a 2 b 1 = 0; or symbolically, we<br />
write in the determinant notation<br />
ij<br />
a1<br />
b1<br />
≡ a 1 b 2 – a 2 b 1 = 0<br />
a 2 b2<br />
Here the scalar a 1 b 2 – a 2 b 1 is said to be the expansion<br />
a1<br />
b1<br />
of the 2 × 2 order determinant having 2<br />
a 2 b2<br />
rows and 2 columns.<br />
Similarly, a determinant of 3 × 3 order can be<br />
expanded as :<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 47 MAY 2010
a<br />
a<br />
a<br />
1<br />
2<br />
3<br />
b<br />
b<br />
b<br />
1<br />
2<br />
3<br />
c<br />
c<br />
c<br />
1<br />
2<br />
3<br />
b2<br />
c2<br />
= a 1<br />
b3<br />
c3<br />
a 2 c2<br />
a 2 b2<br />
– b 1 + c 1<br />
a 3 c3<br />
a 3 b3<br />
= a 1 (b 2 c 3 – b 3 c 2 ) – b 1 (a 2 c 3 – a 3 c 2 ) + c 1 (a 2 b 3 – a 3 b 2 )<br />
= a 1 (b 2 c 3 – b 3 c 2 ) – a 2 (b 1 c 3 – b 3 c 1 ) + a 3 (b 1 c 2 – b 2 c 1 )<br />
= Σ(± a i b j c k )<br />
To every square matrix A = [a ij ] m × n is associated a<br />
number of function called the determinant of A and is<br />
denoted by | A | or det A.<br />
Thus, | A | =<br />
a11<br />
a 21<br />
M<br />
a n1<br />
a12<br />
a 22<br />
M<br />
a n2<br />
...<br />
...<br />
...<br />
a1n<br />
a 2n<br />
M<br />
a nn<br />
If A = [a ij ] n × n , then the matrix obtained from A after<br />
deleting ith row and jth column is called a submatrix<br />
of A. The determinant of this submatrix is called a<br />
minor or a ij .<br />
Sum of products of elements of a row (or column) in<br />
a det with their corresponding cofactors is equal to<br />
the value of the determinant.<br />
n<br />
i.e., ∑<br />
=<br />
i 1<br />
aij<br />
C ij = | A | and ∑<br />
j=<br />
1<br />
n<br />
a C ij = | A |.<br />
(i) If all the elements of any two rows or two columns<br />
of a determinant ate either identical or<br />
proportional, then the determinant is zero.<br />
(ii) If A is a square matrix of order n, then<br />
| kA | = k n | A |.<br />
(iii) If ∆ is determinant of order n and ∆′ is the<br />
determinant obtained from ∆ by replacing the<br />
elements by the corresponding cofactors, then<br />
∆′ = ∆ n–1<br />
(iv) Determinant of a skew-symmetric matrix of odd<br />
order is always zero.<br />
The determinant of a square matrix can be evaluated<br />
by expanding from any row or column.<br />
If A = [a ij ] n × n is a square matrix and C ij is the<br />
cofactor of a ij in A, then the transpose of the matrix<br />
obtained from A after replacing each element by the<br />
corresponding cofactor is called the adjoint of A and<br />
is denoted by adj. A.<br />
Thus, adj. A = [C ij ]′.<br />
Properties of adjoint of a square matrix<br />
(i) If A is a square matrix of order n, then<br />
A . (adj. A) = (adj . A) A = | A | I n .<br />
(ii) If | A | = 0, then A (adj. A) = (adj. A) A = O.<br />
(iii) | adj . A | = | A | n –1 if | A | ≠ 0<br />
(iv) adj. (AB) = (adj. B) (adj. A).<br />
(v) adj. (adj. A) = | A | n – 2 A.<br />
ij<br />
Let A be a square matrix of order n. Then the inverse of<br />
A is given by A –1 1<br />
= adj. A.<br />
| A |<br />
Reversal law : If A, B, C are invertible matrices of same<br />
order, then<br />
(i) (AB) –1 = B –1 A –1<br />
(ii) (ABC) –1 = C –1 B –1 A –1<br />
Criterion of consistency of a system of linear equations<br />
(i) The non-homogeneous system AX = B, B ≠ 0 has<br />
unique solution if | A | ≠ 0 and the unique solution is<br />
given by X = A –1 B.<br />
(ii) Cramer’s Rule : If | A | ≠ 0 and X = (x 1 , x 2 ,..., x n )′<br />
| A<br />
then for each i =1, 2, 3, …, n ; x i =<br />
i |<br />
where<br />
| A |<br />
Ai is the matrix obtained from A by replacing the<br />
ith column with B.<br />
(iii) If | A | = 0 and (adj. A) B = O, then the system<br />
AX = B is consistent and has infinitely many<br />
solutions.<br />
(iv) If | A | = 0 and (adj. A) B ≠ O, then the system<br />
AX = B is inconsistent.<br />
(v) If | A | ≠ 0 then the homogeneous system AX = O<br />
has only null solution or trivial solution<br />
(i.e., x 1 = 0, x 2 = 0, …. x n = 0)<br />
(vi) If | A | = 0, then the system AX = O has non-null<br />
solution.<br />
(i) Area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 )<br />
and (x 3 , y 3 ) is given by<br />
(ii) Three points A(x 1 , y 1 ), B(x 2 , y 2 ) and C(x 3 , y 3 ) are<br />
collinear iff area of ∆ABC = 0.<br />
A square matrix A is called an orthogonal matrix if<br />
AA′ = AA′ = I.<br />
A square matrix A is called unitary if AA θ = A θ A = I<br />
(i) The determinant of a unitary matrix is of modulus<br />
unity.<br />
(ii) If A is a unitary matrix then A′, A , A θ , A –1 are<br />
unitary.<br />
(iii) Product of two unitary matrices is unitary.<br />
Differentiation of Determinants :<br />
Let A = | C 1 C 2 C 3 | is a determinant then<br />
dA = | C′1 C 2 C 3 | + | C 1 C′ 2 C 3 | + | C 1 C 2 C′ 3 |<br />
dx<br />
Same process we have for row.<br />
Thus, to differentiate a determinant, we differentiate one<br />
column (or row) at a time, keeping others unchanged.<br />
1<br />
2<br />
x<br />
x<br />
x<br />
1<br />
2<br />
3<br />
y<br />
y<br />
y<br />
1<br />
2<br />
3<br />
1<br />
1<br />
1<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 48 MAY 2010
a<br />
Based on New Pattern<br />
<strong>IIT</strong>-<strong>JEE</strong> <strong>2011</strong><br />
XtraEdge Test Series # 1<br />
Time : 3 Hours<br />
Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile<br />
motion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current,<br />
Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, Chemical<br />
Kinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics:<br />
Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of<br />
Circle, Function, Limit, Continiuty, Differentiation, Application of Differentiation (Tangent & Normal,<br />
Monotonicity, Maxima & Minima)<br />
Instructions :<br />
Section - I<br />
• Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for<br />
correct answer and -1 mark for wrong answer.<br />
• Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks will be awarded for<br />
correct answer and -1 mark for wrong answer.<br />
• Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer<br />
and -1 mark for wrong answer.<br />
Section - II<br />
• Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly<br />
matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.<br />
PHYSICS<br />
Question 1 to 8 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
1. A target is made of two plates, one of wood and the<br />
other of iron. The thickness of the wooden plate is 4<br />
cm and that of iron plate is 2 cm. A bullet fired<br />
goes through the wood first and then penetrates 1<br />
cm into iron. A similar bullet fired with the same<br />
velocity from opposite direction goes through iron<br />
first and then penetrates 2 cm into wood. If a 1 and<br />
a 2 be the retardations offered to the bullet by wood<br />
and iron plates respectively then -<br />
(A) a 1 = 2a 2 (B) a 2 = 2a 1<br />
(C) a 1 = a 2<br />
(D) Data Insufficient<br />
2. The cone falling with a speed v 0 strikes and<br />
penetrates the block of packing material. The<br />
acceleration of the cone after impact is a = g – cx 2 ,<br />
where c is a positive constant and x is the<br />
penetration distance. If the maximum penetration<br />
depth is x m . Then c equals -<br />
v 0<br />
x<br />
(A)<br />
(C)<br />
2 gx + v<br />
m<br />
2<br />
x m<br />
m<br />
3<br />
2x m<br />
2<br />
0<br />
6gx<br />
− 3v<br />
2<br />
0<br />
(B)<br />
(D)<br />
2gx<br />
− v<br />
m<br />
2<br />
x m<br />
m<br />
3<br />
2x m<br />
2<br />
0<br />
6 gx + 3v<br />
3. A dipole consists of two particles, one of charge Q,<br />
mass m and the other of charge –Q and mass 2m<br />
separated by a distance L. For small oscillations<br />
about the equilibrium position the time after which<br />
the dipole will align itself in the direction of the<br />
uniform field E is -<br />
(A) 2π<br />
(C) 2<br />
π<br />
2m<br />
2mL<br />
3QE<br />
2mL<br />
3QE<br />
–Q<br />
m<br />
(B) π<br />
Q<br />
(D) 4<br />
π<br />
2mL<br />
3QE<br />
2<br />
0<br />
→<br />
E<br />
2mL<br />
3QE<br />
4. In the circuit shown, a potential difference of 60 V<br />
is applied across AB. The potential difference<br />
between the points P and Q is -<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 49<br />
MAY 2010
A<br />
90V<br />
S<br />
2C<br />
B<br />
Q<br />
R<br />
(A) 15 V (B) 30 V (C) 45 V (D) 60 V<br />
C<br />
2C<br />
5. A long straight wire along the z-axis carries a<br />
current I in the negative z direction. The magnetic<br />
vector field → B at a point having coordinates (x, y)<br />
in the z = 0 plane is -<br />
µ ⎛ ⎞<br />
0I<br />
(A) ⎜<br />
yî − xĵ<br />
⎟<br />
µ ⎛ ⎞<br />
0I<br />
(B) ⎜<br />
xî + yĵ<br />
⎟<br />
π<br />
2 2<br />
2<br />
⎝ x + y<br />
2 2<br />
⎠ 2π<br />
⎝ x + y ⎠<br />
µ ⎛ ⎞<br />
0I<br />
(C) ⎜<br />
xĵ−<br />
yî<br />
⎟<br />
µ ⎛ ⎞<br />
0I<br />
(D) ⎜<br />
xî − yĵ<br />
⎟<br />
π<br />
2 2<br />
2<br />
⎝ x + y<br />
2 2<br />
⎠ 2π<br />
⎝ x + y ⎠<br />
6. Two coils, X and Y, are linked such that emf E is<br />
induced in Y when the current in X is changing at<br />
⎛ dI ⎞<br />
the rate I • ⎜ = ⎟ . If a current I 0 is now made to<br />
⎝ dt ⎠<br />
flow through Y, the flux linked with X will be -<br />
•<br />
⎛<br />
(A) EI 0 I (B)<br />
⎟ ⎟ ⎞<br />
⎜ E<br />
I0<br />
⎜ •<br />
⎝ I ⎠<br />
•<br />
(C) (E I)<br />
I 0<br />
(D)<br />
C<br />
I<br />
E<br />
P<br />
•<br />
0 I<br />
7. In the circuit shown, A is joined to B for a long<br />
time, and then A is joined to C. The total heat<br />
produced in R is -<br />
R<br />
Questions 9 to 12 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which MULTIPLE (ONE OR MORE) is correct.<br />
9. Two objects have life times given by t 1 and t 2 . if t is<br />
the life time of an object lying midway between<br />
these two times on the logarithmic scale then -<br />
(A) log 10 (t) = 2<br />
1 [log10 (t 1 ) + log 10 (t 2 )]<br />
t t 2<br />
(B) t =<br />
2<br />
(C) t = t 1 t 2<br />
1 1 ⎛ 1<br />
(D) = t 2 ⎜ +<br />
⎝ t<br />
1 +<br />
1 t 2<br />
1 ⎞<br />
⎟<br />
⎠<br />
10. The position of a particle traveling along x-axis is<br />
given by x t = t 3 – 9t 2 + 6t where x t is in cm and t is<br />
in second Then -<br />
(A) the body comes to rest firstly at (3 – 7 ) s and<br />
then at (3 + 7 ) s<br />
(B) the total displacement of the particle in<br />
traveling from the first zero of velocity to the<br />
second zero of velocity is zero<br />
(C) the total displacement of the particle in<br />
traveling from the first zero of velocity to the<br />
second zero of velocity is – 74 cm.<br />
(D) the particle reverses its velocity at (3 – 7 ) s<br />
and then at (3 + 7 ) s and has a negative<br />
velocity for (3 – 7 ) < t < (3 + 7 )<br />
11. A projectile is fired upward with velocity v 0 at an<br />
angle θ and strikes a point P(x, y) on the roof of the<br />
building (as shown). Then,<br />
y<br />
Roof<br />
(A)<br />
(C)<br />
LE<br />
R<br />
LE<br />
4R<br />
2<br />
2<br />
2<br />
2<br />
+<br />
2L<br />
E<br />
–<br />
2R<br />
(B)<br />
A<br />
LE<br />
2R<br />
2<br />
2<br />
2<br />
LE<br />
(D)<br />
8R 2<br />
8. In a transformer, N P and N S are 1000 and 3000<br />
respectively. If the primary is connected across 80<br />
V A.C., the potential difference across each turn of<br />
the secondary will be -<br />
(A) 240 V<br />
(B) 0.24 V<br />
(C) 0.9 V<br />
(D) 0.08 V<br />
C<br />
B<br />
h<br />
θ<br />
v 0<br />
P(x,y)<br />
(A) the projectile hits the roof in minimum time if<br />
π<br />
θ + α = . 2<br />
(B) the projectile hits the roof in minimum time if<br />
π<br />
θ + α = . 4<br />
(C) the minimum time taken by the projectile to hit<br />
2<br />
2<br />
v0<br />
− v0<br />
− 2gh cos α<br />
the roof is t min =<br />
.<br />
g cosα<br />
(D) the projectile never reaches the roof for<br />
v 0 < 2 gh cosα<br />
α<br />
x<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 50<br />
MAY 2010
12. In the given circuit the point A is 9 V higher than<br />
point B -<br />
A B C D<br />
6V<br />
1Ω<br />
15V<br />
2Ω<br />
24V<br />
1Ω<br />
R<br />
(A) R = 1 Ω<br />
(B) R = 7 Ω<br />
(C) Potential difference between B and D is 30 V<br />
(D) Potential difference between B and C is 15 V<br />
This section contains 2 paragraphs, each has 3<br />
multiple choice questions. (Question 13 to 18) Each<br />
questions has 4 choices (A), (B), (C) and (D) out of<br />
which ONLY ONE is correct.<br />
Passage : I (No. 13 to 15)<br />
In the diagram (given below), the broken lines<br />
represent the paths followed by particles W, X, Y<br />
and Z respectively through the constant field E.<br />
The numbers below the field represent meters.<br />
E<br />
X<br />
Z<br />
Y<br />
W<br />
0 1 2 3 4<br />
13. If all particles started from rest, and all are<br />
positively charged, which particles must have been<br />
acted upon by a force other than that produced by<br />
the electric field –<br />
(A) W and Y (B) X and Z<br />
(C) X, Y and Z (D) W, X, Y and Z<br />
14. If the particles are positively charged, which<br />
particles increased their electric potential energy –<br />
(A) X and Z<br />
(B) Y and Z<br />
(C) W, X, Y and Z<br />
(D) Since the electric field is constant, none of the<br />
particles increased their electric potential<br />
energy<br />
15. Suppose that the field strength E is 10 N/C and<br />
particle Y has a charge of –10 C. When particle<br />
Y is released from rest, it follows the path as<br />
shown and accelerates to a velocity of 10 m/s.<br />
What is the mass of particle Y –<br />
(A) 1 kg<br />
(B) 2 kg<br />
(C) 3 kg<br />
(D) 4 kg<br />
Passage : II (No. 16 to 18)<br />
A set of experiments in the physics lab is designed<br />
to develop understanding of simple electrical circuit<br />
principles for direct current circuits . The student is<br />
given a variety of batteries, resistors, and DC<br />
meters ; and is directed to wire series and parallel<br />
combinations of resistors and batteries making<br />
measurements of the currents and voltage drops<br />
using the ammeters and voltmeters. The student<br />
calculates expected current and voltage values<br />
using Ohm’s law and Kirchoff’s circuit rules and<br />
then checks the results with the meters.<br />
16. Resistors of 4 ohms and 8 ohms are connected in<br />
series. A battery of 6 volts is connected across the<br />
series combination . How much power, in watts, is<br />
consumed in the 8-ohm resistor ?<br />
(A) 0.67 W<br />
(B) 2 W<br />
(C) 12 W<br />
(D) 24 W<br />
17. Two 4-ohm resistors are connected in series and<br />
this pair is connected in parallel with an 8-ohm<br />
resistor. A 12 volt battery is connected across the<br />
ends of this parallel set. What power, in watts, is<br />
consumed in the 8-ohm resistor in this case ?<br />
(A) 0.9 W<br />
(B) 2.0 W<br />
(C) 4.4 W<br />
(D) 18 W<br />
18. A 6-volt battery is connected across a 2- ohm<br />
resistor. What is the heat energy dissipated in the<br />
resistor in 5 minutes ?<br />
(A) 430 joules<br />
(B) 560 joules<br />
(C) 4300 joules<br />
(D) 5400 joules<br />
This section contains 2 questions (Questions 19, 20).<br />
Each question contains statements given in two<br />
columns which have to be matched. Statements (A, B,<br />
C, D) in Column I have to be matched with<br />
statements (P, Q, R, S, T) in Column II. The answers<br />
to these questions have to be appropriately bubbled<br />
as illustrated in the following example. If the correct<br />
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and<br />
D-S, D-T then the correctly bubbled 4 × 5 matrix<br />
should be as follows :<br />
A<br />
B<br />
C<br />
D<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
Mark your response in OMR sheet against the<br />
question number of that question in section-II. + 8<br />
marks will be given for complete correct answer (i.e.<br />
+2 marks for each correct row) and No Negative<br />
marks for wrong answer.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 51<br />
MAY 2010
19. A circular current carrying loop is placed in x-y 2. The potential of the Daniell cell,<br />
(C) 2 × 10 –2 Ms –1 (D) 7 × 10 –3 Ms –1 OA = 2L mol –1<br />
Hence, rate at the start of the reaction is –<br />
plane as shown. A uniform magnetic field B = B 0 kˆ<br />
ZnSO<br />
Zn 4 CuSO 4<br />
Cu was reported by Buckbee,<br />
is present in the region. Match the following :<br />
(1M) (1M )<br />
y<br />
Surdzial, and Metz as<br />
Eº = 1.1028 – 0.641 × 10 –3 T + 0.72 × 10 –5 T 2 ,<br />
x<br />
where T is the celcius temperature. Calculate ∆Sº<br />
for the cell reaction at 25º C -<br />
(A) – 45.32 (B) – 34.52<br />
(C) – 25.43 (D) – 54.23<br />
Column-I<br />
Column-II<br />
(A) Magnetic moment of loop (P) Zero<br />
3. In a hypothetical solid C atoms form CCP lattice<br />
(B) Torque on the loop (Q) Maximum<br />
with A atoms occupying all the Tetrahedral voids<br />
and B atoms occupying all the octahedral voids. A<br />
(C) Potential energy of loop (R) Along +ve<br />
and B atoms are of the appropriate size such that<br />
z-axis<br />
there is no distortion in the CCP lattice. Now<br />
(D) Equilibrium of loop (S) Stable<br />
if a plane is cut (as shown) then the cross<br />
(T) Minimum<br />
section would like –<br />
20. An L-C circuit consists of an inductor with L = 0.09<br />
H and a capacitor of C = 4 × 10 –4 F. The initial<br />
Plane<br />
charge on the capacitor is 5 µC, and the initial<br />
current in the inductor is zero. Match the following:<br />
Column-I<br />
Column-II<br />
(A) Maximum voltage across (P) 8.33 × 10 –4<br />
CCP unit cell<br />
capacitor<br />
(S.I. unit)<br />
(B) Maximum current in the (Q) 3.125 × 10 –8<br />
C B<br />
inductor<br />
(S.I. unit)<br />
A<br />
C C C C C<br />
(C) Maximum energy stored (R) 4.33 × 10 –6<br />
(A) B B B (B) B B B<br />
A<br />
In the inductor<br />
(S.I. unit)<br />
(D) Charge on the capacitor (S) 1.25 × 10 –2<br />
C B C C C C<br />
when current in the inductor (S.I. unit)<br />
has half its maximum value<br />
C C C C C C C<br />
(T) None<br />
A A<br />
(C) B B B (D) B B B<br />
A A<br />
CHEMISTRY<br />
C C C C C C<br />
Question 1 to 8 are multiple choice questions. Each 4. The melting point of RbBr is 682ºC while that of<br />
questions has four choices (A), (B), (C) and (D), out of<br />
NaF is 988ºC. The principal reason that the melting<br />
which ONLY ONE is correct.<br />
point of NaF is much higher than that of RbBr is<br />
that :<br />
1. A reaction follows the given concentration (C) vs<br />
time graph. The rate for this reaction at 20 seconds<br />
(A) the molar mass of NaF is smaller than that of<br />
will be –<br />
RbBr<br />
0.5<br />
(B) the bond in RbBr has more covalent character<br />
0.4<br />
than the bond in NaF<br />
(C) the difference in electronegativity between Rb<br />
0.3<br />
and Br is smaller than the difference between<br />
0.2<br />
Na and F<br />
0.1<br />
(D) the internuclear distance, r c + r a is greater for<br />
RbBr than for NaF<br />
0 20 40 60 80 100<br />
Time/second<br />
5. Following is the graph between (a – x) –1 and time t<br />
for a second order reaction<br />
(A) 4 × 10 –3 Ms –1 (B) 8 × 10 –2 Ms –1<br />
θ = tan –1 (0.5)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 52<br />
MAY 2010
(a–x) –1<br />
A<br />
O time t<br />
(A) 1.25 mol –1 min –1<br />
(B) 0.5 mol L –1 min –1<br />
(C) 0.125 mol L –1 min –1<br />
(D) 1.25 mol L –1 min –1<br />
θ<br />
6. A solution contains Na 2 CO 3 and NaHCO 3 . 10 ml of<br />
the solution requires 2.5 ml of 0.1 M H 2 SO 4 for<br />
neutralization using phenolphthalein as the<br />
indicator. Methyl orange is then added when a<br />
further 2.5 ml of 0.2 M H 2 SO 4 was required. The<br />
amount of Na 2 CO 3 in the g/litre is<br />
(A) 5.3 (B) 4.2 (C) 10.6 (D) 8.4<br />
7. 1 mole mixture of CH 4 & air (containing 80% N 2 ,<br />
20% O 2 by volume) of a composition such that<br />
when underwent combustion gave maximum heat<br />
(assume combustion of only CH 4 ). Then which of<br />
the statements are correct, regarding composition of<br />
initial mixture:<br />
1 2 8<br />
(A) XCH = ,X ,X<br />
4 O =<br />
2<br />
N =<br />
2<br />
11 11 11<br />
3 1 1<br />
(B) XCH = ,X ,X<br />
4 O =<br />
2 N =<br />
2<br />
8 8 2<br />
1 1 2<br />
(C) XCH = ,X , X<br />
4 O =<br />
2 N =<br />
2<br />
6 6 3<br />
(D) Data insufficient<br />
8. Consider the following I st order competing<br />
reactions :<br />
k<br />
X ⎯⎯→<br />
1<br />
k<br />
A + B and Y ⎯⎯→<br />
2 C + D<br />
if 50% of the reaction of X was completed when<br />
96% of the reaction of Y was completed, the ratio<br />
of their rate constants (k 2 / k 1 ) is :<br />
(A) 4.06 (B) 0.215 (C) 1.1 (D) 4.65<br />
Questions 9 to 12 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which MULTIPLE (ONE OR MORE) is correct.<br />
9. Select the correct statements from the following<br />
regarding sols –<br />
(A) Viscosity of lyophilic sols (emulsoid) is much<br />
higher than that of solvent<br />
(B) Surface tension of lyophobic sols (suspensoid)<br />
is usually low<br />
(C) The particles of lyophilic sols always carry a<br />
characteristics charge either positive or<br />
negative<br />
(D) Hydrophobic sols can easily be coagulated by<br />
addition of electrolytes<br />
10. Which of the following is/are correct ?<br />
(A) α-rays are more penetrating than β-rays<br />
(B) α-rays have greater ionizing power than β-rays<br />
(C) β-particles are not present in the nucleus, yet<br />
they are emitted from the nucleus<br />
(D) γ-rays are not emitted simultaneously with α<br />
and β-rays.<br />
11. Choose the correct statement(s) -<br />
(A) At the anode, the species having minimum<br />
reduction potential is formed from the<br />
oxidation of corresponding oxidizable species<br />
(B) In highly alkaline medium, the anodic process<br />
during the electrolytic process is<br />
4OH – → O 2 + 2H 2 O + 4e –<br />
(C) The standard potential of Cl – | AgCl | Ag half–<br />
cell is related to that of Ag + | Ag through the<br />
expression<br />
º<br />
Ag<br />
E + =<br />
/ Ag<br />
º<br />
E<br />
Cl – +<br />
|AgCl|Ag<br />
RT ln Ksp (AgCl)<br />
F<br />
(D) Compounds of active metals (Zn, Na, Mg) are<br />
reducible by H 2 whereas those of noble metals<br />
(Cu, Ag, Au) are not reducible.<br />
12. The co-ordination number of FCC structure for<br />
metals is 12, since -<br />
(A) each atom touches 4 others in same layer, 3 in<br />
layer above and 3 in layer below<br />
(B) each atom touches 4 others in same layer, 4 in<br />
layer above and 4 in layer below<br />
(C) each atom touches 6 others in same layer, 3 in<br />
layer above and 3 in layer below<br />
(D) each atom touches 3 others in same layer, 6 in<br />
layer above and 6 in layer below<br />
This section contains 2 paragraphs, each has 3<br />
multiple choice questions. (Question 13 to 18) Each<br />
questions has 4 choices (A), (B), (C) and (D) out of<br />
which ONLY ONE is correct.<br />
Passage : I (No. 13 to 15)<br />
The cell potential for the unbalanced chemical reaction :<br />
Hg 2<br />
2+<br />
+ NO 3<br />
–<br />
+ H 3 O + ⎯→ Hg 2+ +HNO 2 + H 2 O<br />
is measured under standard state conditions in the<br />
electrochemical cell shown in the accompanying<br />
diagram. The cell voltage is positive: Eº Cell = 0.02V<br />
Eº = 0.02 V<br />
Dish A anode Dish B cathode<br />
Given<br />
NO 3<br />
–<br />
+ 3H 3 O + 2e – ⎯→ HNO 2 + 4H 2 O<br />
Eº = 0.94 V<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 53<br />
MAY 2010
13. Which of the following statements must be true of<br />
the solutions in order for the cell to operate with the<br />
voltage indicated ?<br />
(A) The solution in Dish A must be acidic<br />
(B) The solution in Dish B must be acidic<br />
(C) The solutions in both Dish A and Dish B must<br />
be acidic<br />
(D) No acid may be in either Dish A or Dish B<br />
14. At what pH will the cell potential be zero if the<br />
activity of other components are equal to one ?<br />
(A)<br />
(C)<br />
0.02<br />
2×<br />
0.059<br />
0.04<br />
0.059<br />
(B) –<br />
(D)<br />
0.02<br />
0.059<br />
0.02<br />
0.059<br />
× 3<br />
2<br />
15. How many moles of electrons pass through the<br />
circuit when 0.6 mole of Hg 2+ and 0.30 mole of<br />
HNO 2 are produced in the cell that contains 0.5<br />
2+<br />
mole of Hg 2 and 0.40 mole of NO – 3 at the<br />
beginning of the reaction ?<br />
(A) 0.6 mole (B) 0.8 mole<br />
(C) 0.3 mole (D) 1 mole<br />
Passage : II (No. 16 to 18)<br />
At any fixed temperature, the vapour phase is<br />
always richer in more volatile component as<br />
compared to the solution phase. In other words,<br />
mole fraction of more volatile component is always<br />
greater in vapour phase than in solution phase. We<br />
can also say that vapour phase is relatively richer in<br />
the component whose addition to liquid mixture<br />
results in an increase in total vapour pressure.<br />
16. If 2 moles of A and 3 moles of B are mixed to<br />
form an ideal solution, vapour pressures of A and B<br />
are 120 and 180 mm of Hg respectively, the total<br />
vapour pressure of solution will be<br />
(A) 48 mm Hg (B) 108 mm Hg<br />
(C) 156 mm Hg (D) 15.6 mm Hg<br />
17. From the statement in question 75, the composition<br />
of A and B in the vapour phase when the first traces<br />
of vapours are formed is :<br />
(A) A = 0.407, B = 0.592<br />
(B) A = 0.8, B = 0.1<br />
(C) A = 0.109, B = 0.791<br />
(D) A = 0.307, B = 0.692<br />
18. From the statement in question 75, at what pressure<br />
will the last traces of liquid disappear ?<br />
(A) 100 mm Hg (B) 130 mm Hg<br />
(C) 140 mm Hg (D) 150 mm Hg<br />
This section contains 2 questions (Questions 19, 20).<br />
Each question contains statements given in two<br />
columns which have to be matched. Statements (A, B,<br />
C, D) in Column I have to be matched with<br />
statements (P, Q, R, S, T) in Column II. The answers<br />
to these questions have to be appropriately bubbled<br />
as illustrated in the following example. If the correct<br />
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and<br />
D-S, D-T then the correctly bubbled 4 × 5 matrix<br />
should be as follows :<br />
A<br />
B<br />
C<br />
D<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
Mark your response in OMR sheet against the<br />
question number of that question in section-II. + 8<br />
marks will be given for complete correct answer (i.e.<br />
+2 marks for each correct row) and No Negative<br />
marks for wrong answer.<br />
19. Match the following :<br />
Column-I<br />
Column-II<br />
(A) Intermolecular (P) Ne<br />
H-bonding<br />
(B) Intramolecular (Q) NaCl<br />
H-bonding<br />
(C) Vander Waal's forces (R) H 2 O<br />
OH<br />
(D) Strongest bonding<br />
(S)<br />
CHO<br />
(T) Chloroal<br />
hydrate<br />
20. Match the following :<br />
Column-I<br />
Column-II<br />
(A) PH 3 (P) ≈ 90º or = 90º<br />
(B) H 2 O<br />
(Q) 100º < B.A.
2. If a, b, c, d are the sides of a quadrilateral, then<br />
a + b + c<br />
the minimum value of<br />
2<br />
d<br />
(A) 1 (B) 1/2<br />
(C)1/3 (D) 1/4<br />
3. If the function f(x) = cos |x| – 2ax + b increases<br />
along the entire number scale, the range of<br />
values of a is given by<br />
(A) a ≤ b (B) a = b/2<br />
(C) a ≤ – 1/2 (D) a ≥ – 3/2<br />
⎛ ⎞<br />
4. If φ(x) = 3f ⎜<br />
x 2 ⎟<br />
+ f(3– x 2 ) ∀ x ∈(–3, 4)<br />
⎝ 3 ⎠<br />
where f"(x) > 0 ∀ x ∈ (–3, 4), then φ(x) is<br />
⎛ 3 ⎞<br />
(A) increasing in ⎜ ,4⎟ ⎝ 2 ⎠<br />
⎛ 3 ⎞<br />
(B) decreasing in ⎜−<br />
3 , − ⎟<br />
⎝ 2 ⎠<br />
⎛ 3 ⎞<br />
(C) increasing in ⎜ − , 0 ⎟<br />
⎝ 2 ⎠<br />
⎛ 3 ⎞<br />
(D) decreasing in ⎜0<br />
, ⎟<br />
⎝ 2 ⎠<br />
5. The point in the interval [0, 2π] where f(x) = e x sin x<br />
has maximum slope is<br />
π<br />
π<br />
(A) (B) 4 2<br />
(C) π<br />
2<br />
2<br />
2<br />
is<br />
(D) None of these<br />
a cosA + bcosB + c cos C a + b + c<br />
Q.6 If in ∆ABC,<br />
=<br />
asin B + bsinC + csin A 9R<br />
then the triangle ABC is<br />
(A) isosceles (B) equilateral<br />
(C) right angled (D) None of these<br />
Q.7 The slope of the normal at the point with<br />
abscissa x = – 2 of the graph of the function<br />
f(x) = |x 2 – x| is of form p/q (where p & q are<br />
coprime) then -<br />
(A) p = 4 (B) q = 2<br />
(C) p + q = 6 (D) p – q = – 2<br />
Q.8 In a ∆ABC, b 2 + c 2 = 1999a 2 , then<br />
cot B + cot C<br />
=<br />
cot A<br />
1<br />
1<br />
(A) (B) 999<br />
1999<br />
(C) 999 (D) 1999<br />
Questions 9 to 12 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which MULTIPLE (ONE OR MORE) is correct.<br />
9. For what triplets of real numbers (a, b, c)<br />
with a ≠ 0, the function : f(x) =<br />
⎪⎧<br />
x ; x ≤1<br />
⎨ 2<br />
⎪⎩ ax + bx + c ; otherwise<br />
is continous for all real x :<br />
(A) {(a, 1 – 2a ,a)/a ∈ R, a ≠ 0}<br />
(B) {(2a, 1 – 2a ,0)/a ∈ R, a ≠ 0}<br />
(C) {(a, b, c)/a, b, c ∈ R, a + b + c = 1, a ≠ 0}<br />
(D) {(a, 1 – 2a,c)/ a, c ∈ R, a ≠ 0}<br />
⎧– 2 x < 0<br />
10. Let f(x) = ⎨<br />
⎩x – 2 x ≥ 0<br />
and g(x) = |f(x)| then :<br />
(A) g(x) is continuous for all values of x.<br />
(B) g(x) is differentiable for all value of x<br />
(C) g(x) is not differentibale at x = 0<br />
(D) g(x) is not differentiable at x = 2<br />
⎛ 1 ⎞<br />
11. If 5f(x) + 3f ⎜ ⎟ = x + 2 and y = xf(x) then<br />
⎝ x ⎠<br />
⎛ dy ⎞<br />
⎜ ⎟ is equal to :<br />
⎝ dx ⎠<br />
x=<br />
1<br />
(A) 14 (B) 7/8<br />
(C) 1<br />
(D) None of these<br />
12. If f(x) and g(x) are differentiable function in<br />
[1, 5] and φ(x) = max {f(x), g(x)}, f(x) – g(x) = 0<br />
has exactly one root α in [1, 5] then :<br />
(A) φ(x) continuous and differentiable at all<br />
points in [1, 5]<br />
(B) φ(x) differentiable in [1, 5]<br />
(C) φ(x) necessarily differentiable in [1, 5]– {α}<br />
(D) φ(x) is not differentiable at x = α<br />
This section contains 2 paragraphs, each has 3<br />
multiple choice questions. (Question 13 to 18) Each<br />
questions has 4 choices (A), (B), (C) and (D) out of<br />
which ONLY ONE is correct.<br />
Passage : I (No. 13 to 15)<br />
⎛ x + y ⎞ f (x) + f (y) + b(n – 2)<br />
Let f ⎜ ⎟ =<br />
and<br />
⎝ n ⎠ n<br />
f '(0) = a ; n ∈ N but n ≠ 2.<br />
Let g(x) = f(x) ; x ≥ 0<br />
= 3x + sin 2x ; x < 0<br />
If f(x) & g(x) both are differentiable function, then -<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 55<br />
MAY 2010
13. Range of sgn g(|x|) includes :<br />
(A) 2 (B) 1<br />
(C) –1<br />
(D) all of the above<br />
14. Which one of the following statements doesn't<br />
hold good :<br />
(A) g(x) is many one function<br />
(B) g(x) is invertible<br />
(C) f(x) is invertible<br />
(D) g(|x|) is non differentiable at x = 0<br />
15. Let h(x) = |g(x) –5x|, then wrong statement is :<br />
(A) h(|x|) is a single valued function<br />
(B) |h(|x|)| is always positive<br />
(C) |h(|x|)| is differentiable every where<br />
(D) h(x) is differentiable every where<br />
Passage : II (No. 16 to 18)<br />
Let z denotes the set of integers. Let p be a<br />
prime number and let z 1 ≡ {0, 1}. Let f : z → z<br />
and<br />
g : z → z 1 are two functions defined as follows:<br />
f(n) = p n ; if n ∈ z and<br />
g(n) = 1 ; if n is a perfect square<br />
= 0 ; otherwise<br />
16. g(f(x)) is :<br />
(A) Manyone into<br />
(C) one-one onto<br />
17. f(g(x)) = p has :<br />
(A) no real root<br />
(B) at most one real root<br />
(C) infinitely many roots<br />
(D) exactly two real root<br />
(B) Manyone onto<br />
(D) one-one into<br />
18. Wrong statement about g(f((x)) is :<br />
(A) it is non periodic function<br />
(B) it is neither even nor odd function<br />
(C) it is even function<br />
(D) it is many one function<br />
This section contains 2 questions (Questions 19, 20).<br />
Each question contains statements given in two<br />
columns which have to be matched. Statements (A, B,<br />
C, D) in Column I have to be matched with<br />
statements (P, Q, R, S, T) in Column II. The answers<br />
to these questions have to be appropriately bubbled<br />
as illustrated in the following example. If the correct<br />
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and<br />
D-S, D-T then the correctly bubbled 4 × 5 matrix<br />
should be as follows :<br />
P Q R S T<br />
A P Q R S T<br />
B P Q R S T<br />
C P Q R S T<br />
D P Q R S T<br />
Mark your response in OMR sheet against the<br />
question number of that question in section-II. + 8<br />
marks will be given for complete correct answer (i.e.<br />
+2 marks for each correct row) and No Negative<br />
marks for wrong answer.<br />
19. Match the following :<br />
Column –I<br />
Column –II<br />
(A) f : R → R is defined as (P) 2<br />
⎪⎧<br />
x 2 + kx + 3 for x ≥ 0<br />
f(x) = ⎨ ⎪⎩ 2kx + 3 for x < 0<br />
if f(x) is injective then 'k'<br />
can be equal to<br />
f (x) − 9<br />
(B) If lim = 3 then<br />
x→2<br />
x − 2<br />
(Q) 5<br />
lim f (x) is<br />
x→2<br />
kx<br />
7 + 8<br />
(C) If lim<br />
x →∞ 7 5x<br />
+ 6<br />
does not (R) 9<br />
exists then 'k' can be<br />
(D) Let f(x) & g(x) satisfy the (S) 12<br />
following properties f(3) = 2,<br />
g(3) = 4, g(0) = 3, f ´ (3) = –1,<br />
g´(3) = 0, g´(0) = 2, If<br />
T(x) = f(g(x)) and U(x) = ln(f(x))<br />
then |T´(0) + 6U´3)| can be equal<br />
(T) –1<br />
20. Match the items of column-I with column-II<br />
Column –I<br />
Column –II<br />
⎡ π⎤<br />
(A) Function f : ⎢0,<br />
⎥ → [0, 1]<br />
⎣ 3 ⎦<br />
(P) one-one<br />
defined by f(x) = sin x is function<br />
(B) Function f :(1, ∞) → (1, ∞) (Q) many-one<br />
x + 3<br />
defined by f(x) = is<br />
x −1<br />
function<br />
⎡ π 4π⎤<br />
(C) Function f: ⎢−<br />
, ⎥ →<br />
⎣ 2 3 ⎦<br />
(R) into<br />
[–1, 1] defined by function<br />
f(x) = sin x is<br />
(D) Function f : (2, ∞)→[8, ∞) (S) onto<br />
x 2<br />
defined by f(x) = is<br />
x − 2<br />
function<br />
(T) odd<br />
Function<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 56<br />
MAY 2010
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 57<br />
MAY 2010
Based on New Pattern<br />
<strong>IIT</strong>-<strong>JEE</strong> 2012<br />
XtraEdge Test Series # 1<br />
Time : 3 Hours<br />
Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion,<br />
Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics:<br />
Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of<br />
Circle<br />
Instructions :<br />
Section - I<br />
• Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct<br />
answer and -1 mark for wrong answer.<br />
• Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks will be awarded for correct<br />
answer and -1 mark for wrong answer.<br />
• Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer<br />
and -1 mark for wrong answer.<br />
Section - II<br />
• Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly<br />
matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.<br />
PHYSICS<br />
Question 1 to 8 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
1. If E, m, L and G denote energy, mass, angular<br />
momentum and universal gravitation constant<br />
2<br />
EL<br />
respectively then dimensions of<br />
5 2 will be that<br />
m G<br />
of -<br />
(A) angle (B) length (C) mass (D) time<br />
2. Three vectors → P , → Q and → R are such that | → P | = | → Q |,<br />
| → R | = 2 | → P | and → P + → Q + → R = → 0 . The angles<br />
between → P & → Q , → Q & → R and → P & → R are respectively<br />
(in degrees) -<br />
(A) 90, 135, 135 (B) 90, 45, 45<br />
(C) 45, 90, 90 (D) 45, 135, 135<br />
3. The velocity of a boat in still water in η times less<br />
than the velocity of flow of the river (η > 1). The<br />
angle with the stream direction at which the boat<br />
must move to minimize drifting is -<br />
(A) sin –1 ⎛ 1<br />
⎟ ⎞<br />
⎜<br />
(B) cot –1 ⎛ 1<br />
⎝ η<br />
⎟ ⎞<br />
⎜<br />
⎠<br />
⎝ η ⎠<br />
π<br />
(C) + sin<br />
–1 ⎛ 1 ⎞ π<br />
2 ⎜<br />
⎟ (D) + cot<br />
–1 ⎛ 1 ⎞<br />
⎝ η ⎠ 2 ⎜<br />
⎟<br />
⎝ η ⎠<br />
4. The position of a particle is given by → r = a cos(ωt)<br />
î + a sin (ωt) ĵ + bt kˆ where ω = T<br />
2π and T is time<br />
period for one revolution of the particle following a<br />
helical path. The distance moved by the particle in<br />
one full turn of the helix is -<br />
4π<br />
2 2 2 2π<br />
2 2 2<br />
(A) a + b ω (B) a ω + b<br />
ω ω<br />
(C) ω<br />
π<br />
2 2 2 2<br />
a<br />
π<br />
+ b ω (D) ω<br />
4 2 2 2<br />
a ω + b<br />
5. If the velocity v of a particle moving along a<br />
straight line decreases linearly with its<br />
displacement s from 20 ms –1 to a value approaching<br />
zero at s = 30 m, then acceleration of the particle at<br />
s = 15 m is -<br />
v(in ms –1 )<br />
20<br />
O<br />
(A) 3<br />
2 ms<br />
–2<br />
(C)<br />
20 ms<br />
–2<br />
3<br />
s (in m)<br />
30<br />
(B) – 3<br />
2 ms<br />
–2<br />
20<br />
(D) – ms<br />
–2<br />
3<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 58<br />
MAY 2010
6. A self-propelled vehicle of mass M whose engine<br />
delivers constant power P has an acceleration<br />
P<br />
a = . To increase the velocity of the vehicle<br />
Mv<br />
from v 1 to v 2 , the distance traveled by it (assuming<br />
no friction) is -<br />
3 P<br />
(A) s = (<br />
2<br />
v 2 – 2 M<br />
v 1 ) (B) s = (<br />
2<br />
v 2 – v 2 1 )<br />
M 3P<br />
M<br />
(C) s = (<br />
3<br />
v 2 – 3 3 P<br />
v 1 ) (D) s = (<br />
3<br />
v 2 – v 1<br />
3 )<br />
3P<br />
M<br />
7. A particle starts from rest and traverses a distance l<br />
with uniform acceleration, then moves uniformly<br />
over a further distance 2l and finally comes to rest<br />
after moving a further distance 3l under uniform<br />
retardation. Assuming entire motion to be<br />
rectilinear motion the ratio of average speed over<br />
the journey to the maximum speed on its way is -<br />
1 2 3 4<br />
(A) (B) (C) (D) 5 5 5 5<br />
8. An express elevator can accelerate or decelerate<br />
with values whose magnitudes are limited to 0.4g.<br />
The elevator attains a maximum vertical speed of<br />
400 metre per minute. The minimum time required<br />
by the elevator to start from rest from the 10 th floor<br />
and to stop at the 30 th floor, a distance 100 m apart<br />
is -<br />
(A) 1.67 s<br />
(B) 16.7 s<br />
(C) 167 s<br />
(D) 1670 s<br />
Questions 9 to 12 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which MULTIPLE (ONE OR MORE) is correct.<br />
9. A physical quantity is measured by using an<br />
instrument having a least count l. Then -<br />
(A) error in the measurement of the physical<br />
quantity can exceed the least count l<br />
(B) error in the measurement of the physical<br />
quantity equals the least count l<br />
(C) error in the measurement of the physical<br />
quantity can be less than the least count l<br />
(D) all (A), (B) and (C) are incorrect<br />
10. A bead is free to slide down a smooth wire tightly<br />
stretched between the points P 1 and P 2 on a vertical<br />
circle of radius R. If the bead starts from rest from<br />
P 1 , the highest point on the circle and P 2 lies<br />
anywhere on the circumference of the circle. Then,<br />
P 1<br />
g<br />
θ<br />
P 2<br />
R<br />
(A) time taken by bead to go from P 1 to P 2 is<br />
dependent on position of P 2 and equals<br />
R<br />
2 cos θ<br />
g<br />
(B) time taken by bead to go from P 1 to P 2 is<br />
R<br />
independent of position of P 2 and equals 2<br />
g<br />
(C) acceleration of bead along the wire is g cos θ<br />
(D) velocity of bead when it arrives at P 2 is<br />
2 gR cos θ<br />
11. A body is moving along a straight line. Its distance<br />
x t from a point on its path at a time t after passing<br />
that point is given by x t = 8t 2 – 3t 3 , where x t is in<br />
metre and t in second.<br />
(A) Average speed during the interval t = 0 s to<br />
t = 4 s is 20.21 ms –1<br />
(B) Average velocity during the interval t = 0 s to<br />
t = 4 s is – 16 ms –1<br />
16<br />
(C) The body starts from rest and at t = s it 9<br />
reverses its direction of motion at x t = 8.43 m<br />
from the start<br />
(D) It has an acceleration of –56 ms –2 at t = 4s<br />
12. A particle is acted upon by a force of constant<br />
magnitude which is always perpendicular to the<br />
velocity of the particle. The motion of the particle<br />
takes place in a plane. It follows that -<br />
(A) its velocity is constant<br />
(B) its acceleration is constant<br />
(C) its kinetic energy is constant<br />
(D) it moves in a circular path<br />
This section contains 2 paragraphs, each has 3<br />
multiple choice questions. (Question 13 to 18) Each<br />
questions has 4 choices (A), (B), (C) and (D) out of<br />
which ONLY ONE is correct.<br />
Passage : I (No. 13 to 15)<br />
A student performs an experiment to determine<br />
how the range of a ball depends on the velocity<br />
with which it is released. The “range” is the<br />
distance between where the ball lands and where it<br />
was released, assuming it lands at the same height<br />
from which it was released.<br />
In each trial, the student uses the same baseball, and<br />
launches it at the same angle. Table I shows the<br />
experimental results<br />
Table 1<br />
Trial Launch speed (m/s) Range (m)<br />
1<br />
2<br />
3<br />
4<br />
10<br />
20<br />
30<br />
40<br />
8.0<br />
31.8<br />
70.7<br />
122.5<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 59<br />
MAY 2010
Based on this data, the student then hypothesizes<br />
that the range, R, depends on the initial speed, v 0 ,<br />
according to the following equation : R = Cv 0 n ,<br />
where C is a constant, and n is another constant.<br />
13. Based on this data, the best guess for the value of n<br />
is –<br />
(A) 1/2 (B) 1 (C) 2 (D) 3<br />
14. The student speculates that the constant C depends<br />
on :<br />
I. The angle at which the ball was launched<br />
II. The ball’s mass<br />
III. The ball’s diameter<br />
If we neglect air resistance, then C actually depends<br />
on –<br />
(A) I only<br />
(B) I and II<br />
(C) I and III (D) I, II and III<br />
15. The student performs another trial in which the ball<br />
is launched at speed 5.0 m/s. Its range is<br />
approximately –<br />
(A) 1.0 meters (B) 2.0 meters<br />
(C) 3.0 meters (D) 4.0 meters<br />
Passage : II (No. 16 to 18)<br />
<br />
<br />
When an airplane flies, its total velocity with<br />
respect to the ground is :<br />
v total = v plane + v wind<br />
Where v plane denotes the plane's velocity through<br />
motionless air, and v wind denotes the wind's<br />
velocity. Crucially, all the quantities in this<br />
equations are vectors. The magnitude of a velocity<br />
vector is often called the "speed".<br />
Consider an airplane whose speed through<br />
motionless air is 100m/s. To reach its destination,<br />
the plane must fly east.<br />
The "heading" of a plane of the direction in which<br />
the nose of the plane points. So, it is the direction in<br />
which the engines propel the plane.<br />
16. If the plane has an eastward heading, and a 20 m/s<br />
wind blows towards the southwest, then the plane's<br />
speed is :<br />
(A) 80 m/s<br />
(B) more than 80 m/s but less than 100 m/s<br />
(C) 100 m/s<br />
(D) more than 100 m/s<br />
17. The pilot maintains an eastward heading while a<br />
20 m/s wind blows northward. The plane's velocity<br />
is deflected from due east by what angle ?<br />
(A) sin –1 (1/5) (B) cos –1 (1/5)<br />
(C) tan –1 (1/5) (D) None of these<br />
18. Because the 20 m/s northward wind persists, the<br />
pilot adjusts the heading so that the plane's total<br />
velocity is eastward. By what angle does the new<br />
heading differ form due east ?<br />
(A) sin –1 1/5 (B) cos –1 1/5<br />
(C) tan –1 1/5 (D) None of these<br />
This section contains 2 questions (Questions 19, 20).<br />
Each question contains statements given in two<br />
columns which have to be matched. Statements (A, B,<br />
C, D) in Column I have to be matched with<br />
statements (P, Q, R, S, T) in Column II. The answers<br />
to these questions have to be appropriately bubbled<br />
as illustrated in the following example. If the correct<br />
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and<br />
D-S, D-T then the correctly bubbled 4 × 5 matrix<br />
should be as follows :<br />
A<br />
B<br />
C<br />
D<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
Mark your response in OMR sheet against the<br />
question number of that question in section-II. + 8<br />
marks will be given for complete correct answer (i.e.<br />
+2 marks for each correct row) and No Negative<br />
marks for wrong answer.<br />
19. Column-I Column-II<br />
→ →<br />
. b<br />
(A) a (P) Area of the<br />
Parallelogram<br />
formed<br />
(B) L<br />
(Q) Area below y-x<br />
graph within the<br />
limits given<br />
(C)<br />
→ →<br />
a × b<br />
(R) Projection of<br />
→ →<br />
b on a<br />
x<br />
(D)<br />
∫ f ydx<br />
x t<br />
(S) Aerial velocity<br />
(T) Zero for → a ⊥ → b<br />
20. Column-I Column-II<br />
(A) Horizontal range is (P) π/2<br />
maximum, when the<br />
projection is vertical<br />
(B) Vertical height is (Q) π<br />
maximum, when the<br />
angle of projection is<br />
(C) x and y coordinates (R) π/4<br />
changes in projectile<br />
motion with<br />
(D) The magnitude of<br />
velocity with the<br />
time in projection<br />
(S) Time<br />
(T) First decrease then<br />
increase<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 60<br />
MAY 2010
CHEMISTRY<br />
Question 1 to 8 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
1. A 0.518 g sample of lime stone is dissolved in HCl<br />
and then the calcium is precipitated as CaC 2 O 4 .<br />
After filtering and washing the precipitate, it<br />
requires 40 ml of 0.250 N KMnO 4 , solution<br />
acidified with H 2 SO 4 to titrate it as,<br />
MnO 4 – + H + + C 2 O 4 –2 → Mn +2 + CO 2 + 2H 2 O<br />
The percentage of CaO in the sample is<br />
(A) 54.0% (B) 27.1% (C) 42% (D) 84%<br />
2. CO 2 is a gas but SiO 2 is a solid, because –<br />
(A) CO 2 has smaller molecular size but SiO 2 has<br />
greater molecular size<br />
(B) In CO 2 pπ – pπ bond exists between C and O<br />
(C) In SiO 2 , pπ – pπ bond exists between Si and O<br />
(D) CO 2 is a discrete molecule where weak<br />
molecular force of attraction exist, while SiO 2 is<br />
a large polymeric molecule<br />
3. In BrF 3 molecule, the lone pairs occupy equatorial<br />
position to minimize –<br />
(A) lone pair-bond pair repulsion only<br />
(B) bond pair-bond pair repulsion only<br />
(C) lone pair-lone pair repulsion and lone pair-bond<br />
pair repulsion<br />
(D) lone pair-lone pair repulsion only<br />
4. Which is/are correct statements ?<br />
(A) A solute will dissolve in water if hydration<br />
energy is greater than lattice energy<br />
(B) If the anion is large compared to the cation, the<br />
lattice energy will remain almost constant<br />
(C) solubility of II A hydroxide is in order<br />
Be(OH) 2 < Mg(OH) 2 < Ca(OH) 2 < Sr(OH) 2<br />
(D) none is correct<br />
5. Sulphur trioxide is prepared by the following two<br />
reactions.<br />
S 8 (s) + 8O 2 (g) → 8SO 2 (g)<br />
2SO 2 (g) + O 2 (g) → 2SO 3 (g)<br />
How many grams of SO 3 are produced from 1 mol<br />
of S 8 ?<br />
(A) 1280.0 (B) 640.0<br />
(C) 960.0 (D) 320.0<br />
6. Which of the following statements is correct ?<br />
(A) The magnitude of the second electron affinity<br />
of Sulphur is greater than that of Oxygen<br />
(B) The magnitude of the second electron affinity<br />
of Sulphur is less than that of Oxygen<br />
(C) The first electron affinities of Bromine and<br />
Iodine are approximately the same<br />
(D) The first electron affinity of Fluorine is greater<br />
than that of Chlorine<br />
7. Iron forms two oxides, in first oxide 56 gram. Iron<br />
is found to be combined with 16 gram oxygen and<br />
in second oxide 112 gram. Iron is found to be<br />
combined with 48 gram oxygen. This data satisfy<br />
the law of -<br />
(A) Conservation of mass<br />
(B) Reciprocal proportion<br />
(C) Multiple proportion<br />
(D) Combining volume<br />
8. Lattice energy of BeCO 3<br />
(I) , MgCO 3<br />
(II) and<br />
CaCO 3<br />
(III) are in the order -<br />
(A) I > II > III (B) I < II < III<br />
(C) I < III < II (D) II < I < III<br />
Questions 9 to 12 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which MULTIPLE (ONE OR MORE) is correct.<br />
9. Which of the following statements is/are correct ?<br />
(A) Group 12(IIB) elements do not show<br />
characteristic properties of transition metals<br />
(B) Among transition elements, tungsten has the<br />
highest melting point<br />
(C) Among transition elements, group 3 (IIIB)<br />
elements have lowest densities<br />
(D) Transition metals are more electropositive than<br />
alkaline earth metals.<br />
10. 11.2 g of mixture of MCl (volatile) and NaCl gave<br />
28.7 g of white ppt with excess of AgNO 3 solution.<br />
11.2 g of same mixture on heating gave a gas that<br />
on passing into AgNO 3 solution gave 14.35 g of<br />
white ppt. Hence ?<br />
(A) Ionic mass of M + is 18<br />
(B) Mixture has equal mol fraction of MCl and<br />
NaCl<br />
(C) MCl and NaCl are in 1 : 2 molar ratio<br />
(D) Ionic mass of M + is 10<br />
11. Specify the coordination geometry around the<br />
hybridization of N and B atoms in a 1 : 1 complex<br />
of BF 3 and NH 3 -<br />
(A) N : tetrahedral, sp 3 ;B : tetrahedral, sp 3<br />
(B) N : pyramidal, sp 3 ; B : pyramidal, sp 3<br />
(C) N : pyramidal, sp 3 ; B : planar, sp 2<br />
(D) N : pyramidal, sp 3 ; B : tetrahedral, sp 3<br />
12. The radii of two of the first four Bohr orbits of the<br />
hydrogen atom are in the ratio 1 : 4. The energy<br />
difference between them may be :<br />
(A) Either 12.09 eV or 3.4 eV<br />
(B) Either 2.55 eV or 10.2 eV<br />
(C) Either 13.6 eV or 3.4 eV<br />
(D) Either 3.4 eV or 0.85 eV<br />
This section contains 2 paragraphs, each has 3<br />
multiple choice questions. (Question 13 to 18) Each<br />
questions has 4 choices (A), (B), (C) and (D) out of<br />
which ONLY ONE is correct.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 61<br />
MAY 2010
Passage : I (No. 13 to 15)<br />
The radius of the nucleus of an atom can be<br />
approximately determined as,<br />
r nu = (1.4×10 –13 )A 1/3<br />
where A is mass number of the atoms and M is the<br />
charge of the electron = 4.8 × 10 –10 esu.<br />
The mass of α-particle = 4 × mass of H-atom<br />
10<br />
mass of hydrogen atom = ×10 –24 gm.<br />
6<br />
Consider during collision kinetic energy of<br />
α-particle just equal to coulombic force of<br />
repulsion.<br />
The mass number of Au = 197<br />
The mass number of He = 4<br />
The atomic number of Au = 79<br />
Given : (4) 1/3 = 1.59 and (197) 1/3 = 5.82<br />
2×<br />
79×<br />
(4.8)<br />
= 351<br />
10.374<br />
3 .51× 3 = 3.245<br />
Plank’s constant, h = 6.625 × 10 –34 JS<br />
13. What is the distance between the α-particle and Au<br />
nucleus during the collision –<br />
(A) 10.374 × 10 –13 cm (B) 10.374 Å<br />
(C) 10.374 × 10 –10 cm (D) 10.374 nm<br />
14. What should be the minimum velocity of the<br />
α-particle to strike the nucleus of 79 Au 197 ?<br />
(A) 3.245 × 10 8 m/s (B) 3.245 × 10 9 m/s<br />
(C) 3.245 × 10 5 m/s (D) 3.245 × 10 7 m/s<br />
15. What is the de-broglie’s wave length associated<br />
with a α-particle while it is moving to colloide with<br />
the Au nucleus ?<br />
6.625× 6<br />
(A) ×10 25 6 .625× 6<br />
m (B) ×10 –15 m<br />
4×<br />
3.245<br />
3.245<br />
6.625× 6<br />
(C) ×10 –15 6.625× 6<br />
m (D) ×10 –15 m<br />
4×<br />
3.245<br />
5×<br />
3.245<br />
Passage : II (No. 16 to 18)<br />
When we use the concept that one mole of one<br />
substance contains the same number of elementary<br />
entities as one mole of any other substance we don't<br />
actually need to know what that number is.<br />
Sometimes however we will need to work with the<br />
actual number of elementary entities in a mole of<br />
substance. This number is called Avogadro's<br />
number.<br />
N A = 6.022137 × 10 23 mol –1<br />
The unit mol –1 'which we read as' per mole signifies<br />
that a collection of N A molecular level entities is<br />
equivalent to one mole at the macroscopic level.<br />
For example a mole of carbon contains 6.02 × 10 23<br />
atoms of C. A mole of oxygen contains 6.02 × 10 23<br />
molecules of O 2 .<br />
2<br />
16. The number of atoms present in 8 g of ozone is<br />
(A) N A<br />
(B) 3N A<br />
(C)<br />
N A<br />
6<br />
(D)<br />
N A<br />
17. Which of the following is a reasonable value for the<br />
number of atoms in 1.00 g of Helium ?<br />
(A) 0.25 (B) 4.0<br />
(C) 4.1 × 10 –23 (D) 1.5 × 10 23<br />
18. How many years it would take to spend Avogadro's<br />
number of rupees at the rate of 10 Lac rupees per<br />
second ?<br />
(A) 1.9 × 10 10 years (B) 1.6 × 10 10 years<br />
(C) 5 × 10 6 years (D) 6.4 × 10 5 years<br />
This section contains 2 questions (Questions 19, 20).<br />
Each question contains statements given in two<br />
columns which have to be matched. Statements (A, B,<br />
C, D) in Column I have to be matched with<br />
statements (P, Q, R, S, T) in Column II. The answers<br />
to these questions have to be appropriately bubbled<br />
as illustrated in the following example. If the correct<br />
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and<br />
D-S, D-T then the correctly bubbled 4 × 5 matrix<br />
should be as follows :<br />
A<br />
B<br />
C<br />
D<br />
2<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
Mark your response in OMR sheet against the<br />
question number of that question in section-II. + 8<br />
marks will be given for complete correct answer (i.e.<br />
+2 marks for each correct row) and No Negative<br />
marks for wrong answer.<br />
19. Match the following :<br />
Column-I<br />
Column-II<br />
–<br />
(A) ICl 2 (P) Linear<br />
+<br />
(B) BrF 2 (Q) Pyramidal<br />
–<br />
(C) ClF 4 (R) Tetrahedral<br />
–<br />
(D) AlCl 4 (S) Square planar<br />
(T) Angular<br />
20. Match the following :<br />
Column-I<br />
Column-II<br />
(A) 0.5 mole of SO 2 (g) (P) Occupy 11.2L<br />
at STP<br />
(B) 1g of H 2 (g) (Q) Weighs 24g<br />
(C) 0.5 mole of O 3 (g) (R) Total no. of<br />
atoms = 1.5N A<br />
(D) 1g molecule of O 2 (g) (S) Weighs 32g<br />
(T) Total no. of atoms<br />
≈ 9 × 10 23<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 62<br />
MAY 2010
MATHEMATICS<br />
Question 1 to 8 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
1. 3 cosec 20° – sec 20° is equal to :<br />
(A) 1 (B) 2<br />
(C) 4<br />
(D) none of these<br />
2. If cos 20° – sin 20° = p then cos 40° is equal to-<br />
(A)<br />
(C)<br />
2<br />
− p 2 − p (B) p 2 − p<br />
2<br />
p + 2 − p (D) none of these<br />
3. The expression<br />
⎡ 4⎛<br />
3π<br />
⎞ 4 ⎤<br />
3⎢sin<br />
⎜ – α⎟ + sin (3π<br />
– α)<br />
⎥<br />
⎣ ⎝ 2 ⎠<br />
⎦<br />
⎡ 6⎛<br />
π ⎞<br />
– 2⎢sin<br />
⎜ + α⎟ + sin<br />
⎣ ⎝ 2 ⎠<br />
is equal to -<br />
(A) 0 (B) 1<br />
(C) 2 (D) 5<br />
2<br />
6<br />
⎤<br />
(5π<br />
– α)<br />
⎥<br />
⎦<br />
4.<br />
π tan α<br />
If α + β = and β + γ = α, then 2 tanβ + 2 tan γ<br />
(A) 0 (B) 1<br />
(C) 2 (D) 3<br />
5. If 1< x < 2 , then number of solutions of the equation<br />
tan –1 (x – 1) + tan –1 x + tan –1 (x + 1) = tan –1 3x, is/are<br />
(A) 0 (B) 1<br />
(C) 2 (D) 3<br />
6. In a triangle ABC if BC =1 and AC = 2. Then the<br />
maximum possible value of angle A is-<br />
(A) π/6 (B) π/4<br />
(C) π/3 (D) π/2<br />
7. If α ∈<br />
⎛ 3π<br />
⎞<br />
⎜− , − π⎟ ,<br />
⎝ 2 ⎠<br />
then the value of<br />
tan –1 (cot α) – cot –1 (tan α) + sin –1 (sin α) + cos –1 (cos α) is<br />
equal to -<br />
(A) 2π + α<br />
(B) π + α<br />
(C) 0<br />
(D) π – α<br />
=<br />
8. Solution of the equation<br />
3 sin –1 ⎛ 2x ⎞ ⎛ 2<br />
⎜ ⎟<br />
⎝1+ x<br />
2 – 4 cos –1<br />
⎠<br />
⎟ ⎟ ⎞<br />
⎜1−<br />
x<br />
+<br />
⎜ 2<br />
⎝1+<br />
x ⎠<br />
2 tan –1 ⎛ 2x ⎞ π<br />
⎜ ⎟<br />
⎝1− x<br />
2 = is -<br />
⎠ 3<br />
(A) x = 3 (B) x =<br />
3<br />
(C) x = 1 (D) x = 0<br />
Questions 9 to 12 are multiple choice questions. Each<br />
questions has four choices (A), (B), (C) and (D), out of<br />
which MULTIPLE (ONE OR MORE) is correct.<br />
9. If sin θ + sin φ = a and cos θ + cos φ = b , then -<br />
⎛ θ – φ ⎞ 1<br />
2<br />
(A) cos⎜<br />
⎟ = ± (a<br />
2 + b )<br />
⎝ 2 ⎠ 2<br />
⎛ θ – φ ⎞ 1 2 2<br />
(B) cos⎜<br />
⎟ = ± (a – b )<br />
⎝ 2 ⎠ 2<br />
⎛ θ – φ ⎞<br />
(C) tan ⎜ ⎟ = ±<br />
⎝ 2 ⎠<br />
(D) cos (θ – φ) =<br />
a<br />
⎛<br />
2<br />
⎜<br />
4 – a – b<br />
2 2<br />
⎝ a + b<br />
2 +<br />
b<br />
2<br />
2<br />
– 2<br />
10. Let y = sin x . sin (60 º + x) . sin(60 º - x). Then for<br />
all real x -<br />
(A) the minimum value of y is 2<br />
(B) the maximum value of y is 1<br />
(C) y ≤ 1/4<br />
(D) y ≥ –1/4<br />
11. If cosec –1 x = sin –1 (1/x), then x may be-<br />
(A) 1 (B) –1/2 (C) 3/2 (D) – 3/2<br />
12. If in a triangle ABC, θ is the angle determined by<br />
cos θ = (a – b)/c, then<br />
( a + b)sin θ A − B<br />
(A)<br />
= cos<br />
2 ab<br />
2<br />
( a + b)sin θ A + B<br />
(B)<br />
= cos<br />
2 ab<br />
2<br />
csin<br />
θ A − B<br />
(C) = cos<br />
2 ab 2<br />
csin<br />
θ A + B<br />
(D) = cos<br />
2 ab 2<br />
2<br />
1<br />
⎞<br />
⎟<br />
⎠<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 63<br />
MAY 2010
This section contains 2 paragraphs, each has 3<br />
multiple choice questions. (Question 13 to 18) Each<br />
questions has 4 choices (A), (B), (C) and (D) out of<br />
which ONLY ONE is correct.<br />
Passage : I (No. 13 to 15)<br />
f(x) = sin {cot –1 (x + 1)} – cos (tan –1 x)<br />
a = cos tan –1 sin cot –1 x<br />
b = cos (2 cos –1 x + sin –1 x)<br />
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and<br />
D-S, D-T then the correctly bubbled 4 × 5 matrix<br />
should be as follows :<br />
A<br />
B<br />
C<br />
D<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
P Q R S T<br />
13. The value of x for which f(x) = 0 is<br />
(A) – 1/2 (B) 0<br />
(C) 1/2 (D) 1<br />
14. If f(x) = 0 then a 2 is equal to<br />
(A) 1/2 (B) 2/3<br />
(C) 5/9 (D) 9/5<br />
15. If a 2 = 26/51, then b 2 is equal to<br />
(A) 1/25 (B) 24/25<br />
(C) 25/26 (D) 50/51<br />
Passage : II (No. 16 to 18)<br />
In a ∆ABC, if cos A. cos B. cos C =<br />
3 + 3<br />
sin A. sin B. sin C = , then<br />
8<br />
16. Value of tan A + tan B + tan C is<br />
3 + 3<br />
3 + 4<br />
(A)<br />
(B)<br />
3 −1<br />
3 −1<br />
(C)<br />
6 −<br />
3<br />
3 −1<br />
(D) None of these<br />
17. Value of Σ tan A .tan B =<br />
(A) 5 – 4 3 (B) 5 + 4 3<br />
(C) 6 + 4 3 (D) 6 – 4 3<br />
18. Value of tan A, tan B and tan C are<br />
(A) 1, 3 , 2 (B) 1, 3 , 2<br />
(C) 1, 2, 3 (D) None of these<br />
3 −1<br />
8<br />
Mark your response in OMR sheet against the<br />
question number of that question in section-II. + 8<br />
marks will be given for complete correct answer (i.e.<br />
+2 marks for each correct row) and No Negative<br />
marks for wrong answer.<br />
19. Match the items of Column I with the items of<br />
Column II. The principal value of<br />
Column-I<br />
Column-II<br />
− ⎛ 5π<br />
⎞<br />
(A) sin 1 17π<br />
⎜sin<br />
⎟ is (P)<br />
⎝ 6 ⎠ 20<br />
− ⎛ 7π<br />
⎞<br />
(B) cos 1 1<br />
⎜−<br />
sin ⎟ is (Q)<br />
⎝ 6 ⎠ 2<br />
⎧<br />
(C)<br />
− ⎛ 15π<br />
⎞⎫<br />
π<br />
1<br />
cos⎨tan<br />
⎜ tan ⎟⎬<br />
is (R)<br />
⎩ ⎝ 4 ⎠⎭<br />
3<br />
− ⎛<br />
(D) cos 1<br />
⎜<br />
⎝<br />
1 ⎛ 9π<br />
9π<br />
⎞⎞<br />
π<br />
⎜cos<br />
− sin ⎟<br />
⎟ is (S)<br />
2 ⎝ 10 10 ⎠⎠<br />
6<br />
(T) 2<br />
π<br />
20. Column-I Column-II<br />
(A) If in a ∆ABC the angles are<br />
5π<br />
(P) 12<br />
in AP and b : c = 3 : 2<br />
then ∠A =<br />
(B) If the length of the side of a<br />
2π<br />
(Q)<br />
3<br />
triangle are 3, 5, 7 then largest<br />
angle of the triangle is<br />
(C) If the sides of a triangle are<br />
π<br />
(R) 6<br />
This section contains 2 questions (Questions 19, 20).<br />
Each question contains statements given in two<br />
columns which have to be matched. Statements (A, B,<br />
C, D) in Column I have to be matched with<br />
statements (P, Q, R, S, T) in Column II. The answers<br />
to these questions have to be appropriately bubbled<br />
as illustrated in the following example. If the correct<br />
in ratio 2 : 6 : ( 3 + 1) then<br />
largest angle of the triangle is<br />
(D) If in a triangle ABC, b = 3 , (S) π<br />
c = 1 and B – C = 90º then<br />
∠A =<br />
(T) 3π/5<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 64<br />
MAY 2010
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 65<br />
MAY 2010
<strong>IIT</strong>-<strong>JEE</strong> 2010<br />
PAPER-I (PAPER & SOLUTION)<br />
Time : 3 Hours Total Marks : 243<br />
Instructions :<br />
• The question paper consists of 3 parts (chemistry, Mathematics and Physics). Each part consists of four sections.<br />
• For each question in Section I, you will be awarded 3 marks if you have darkened only the bubble corresponding to<br />
the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be<br />
awarded.<br />
• For each question in Section II, you will be awarded 3 marks if you darkem only the bubble corresponding to the<br />
correct answer and zero mark if no bubbles are darkened. Partial marks will be awarded for partially correct<br />
answers. No negative marks will be awarded in this section.<br />
• For each question in Section III, you will be awarded 3 marks if you darken only the bubble corresponding to the<br />
correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be<br />
awarded.<br />
• For each question in Section IV, you will be awarded 3 marks if you darken the bubble corresponding to the<br />
correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for in this section.<br />
CHEMISTRY<br />
SECTION – I<br />
Single Correct Choice Type<br />
This section contains 8 multiple choice questions. Each<br />
question has 4 choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
1. The bond energy (in kcal mol –1 ) of a C–C single<br />
bond is approximately<br />
(A) 1 (B) 10<br />
(C) 100 (D) 1000<br />
Ans. [C]<br />
Sol. Value is 82.6 kcal/mol.<br />
2. The species which by definition has ZERO<br />
standard molar enthalpy of formation at 298 K is<br />
(A) Br 2 (g)<br />
(B) Cl 2 (g)<br />
(C) H 2 O (g)<br />
(D) CH 4 (g)<br />
Ans. [B]<br />
Sol. Because standard state of Cl 2 is gas.<br />
3. The ionization isomer of [Cr(H 2 O) 4 Cl(NO 2 )] Cl<br />
is<br />
(A) [Cr(H 2 O) 4 (O 2 N)] Cl 2<br />
(B) [Cr(H 2 O) 4 Cl 2 ](NO 2 )<br />
(C) [Cr(H 2 O) 4 Cl(ONO)]Cl<br />
(D)<br />
[Cr(H 2 O) 4 Cl 2 (NO 2 )]. H 2 O<br />
Ans. [B]<br />
Sol.<br />
Ionization isomer have different ions in solution<br />
4. The correct structure of ethylenediaminetetraacetic<br />
acid (EDTA) is -<br />
HOOC–CH 2<br />
CH 2 –COOH<br />
Ans.<br />
Sol.<br />
(A)<br />
HOOC–CH 2<br />
(B)<br />
HOOC<br />
HOOC<br />
HOOC–CH 2<br />
(C)<br />
HOOC–CH 2<br />
(D)<br />
HOOC–CH 2<br />
N–CH=CH–N<br />
N–CH 2 –CH 2 –N<br />
H<br />
N–CH 2 –CH 2 –N<br />
COOH<br />
CH 2<br />
N–CH–CH–N<br />
HOOC<br />
[C]<br />
Structure is<br />
HOOC–CH 2<br />
HOOC–CH 2<br />
CH 2<br />
CH 2 –COOH<br />
COOH<br />
COOH<br />
CH 2 –COOH<br />
H<br />
N–CH 2 –CH 2 –N<br />
CH 2 –COOH<br />
CH 2 –COOH<br />
CH 2 –COOH<br />
CH 2 –COOH<br />
5. The synthesis of 3-octyne is achieved by adding a<br />
bromoalkane into a mixture of sodium amide and<br />
an alkyne. The bromoalkane and alkyne<br />
respectively are<br />
(A) BrCH 2 CH 2 CH 2 CH 2 CH 3 and CH 3 CH 2 C≡CH<br />
(B) BrCH 2 CH 2 CH 3 and CH 3 CH 2 CH 2 C≡CH<br />
(C) BrCH 2 CH 2 CH 2 CH 2 CH 3 and CH 3 C≡CH<br />
(D) BrCH 2 CH 2 CH 2 CH 3 and CH 3 CH 2 C ≡CH<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 66<br />
MAY 2010
Ans.<br />
[D]<br />
Sol. (i) CH 3 –CH 2 –C≡CH<br />
NaNH 2<br />
CH 3 CH 2 C≡<br />
C –<br />
C –<br />
+<br />
Na<br />
(ii) CH 3 –CH 2 –CH 2 –CH 2 –Br + CH 3 CH 2 –C≡<br />
+<br />
Na<br />
––→ CH 3 –CH 2 –CH 2 –CH 2 –C ≡C–CH 2 –CH 3<br />
6. The correct statement about the following<br />
disaccharide is<br />
CH 2 OH<br />
H O<br />
CH 2 OH O H<br />
H<br />
H (a)<br />
OH H<br />
(b) OH<br />
OCH 2 CH 2 O<br />
H<br />
OH<br />
CH 2 OH<br />
H OH<br />
OH H<br />
(A) Ring (a) is pyranose with α-glycosidic link<br />
(B) Ring (a) is furanose with α-glycosidic link<br />
(C) Ring (b) is furanose with α-glycosidic link<br />
(D) Ring (b) is Pyranose with β-glycosidic link<br />
Ans. [A]<br />
Sol. Ring (a) is pyranose with α-glycosidic link<br />
7. Plots showing the variation of the rate constant<br />
(k) with temperature (T) are given below. The<br />
plot that follows Arrhenius equation is -<br />
Ans.<br />
Sol.<br />
(A)<br />
k<br />
(C)<br />
k<br />
T<br />
T<br />
[A]<br />
k = Ae –Ea/RT ; k ∝ e<br />
k<br />
T<br />
– E a / RT<br />
(B)<br />
k<br />
(D)<br />
k<br />
8. In the reaction OCH 3 HBr the<br />
products are<br />
(A) Br<br />
OCH 3 and H 2<br />
(B)<br />
Br and CH 3 Br<br />
T<br />
T<br />
Ans.<br />
Sol.<br />
(C)<br />
(D)<br />
[D]<br />
Br and CH 3 OH<br />
OH and CH 3 Br<br />
OCH 3<br />
HBr<br />
SECTION – II<br />
Multiple Correct Choice Type<br />
OH + CH 3 Br<br />
This section contains 5 multiple choice questions. Each<br />
questions has 4 choices (A), (B), (C) and (D), out of<br />
which ONE OR MORE is/are correct.<br />
9. In the reaction<br />
intermediate(s) is (are)<br />
(A)<br />
O<br />
Br<br />
O<br />
(C)<br />
Br<br />
Ans. [A, B, C]<br />
OH<br />
Br<br />
Sol. NaOH(aq)/Br 2<br />
Br<br />
OH<br />
(B)<br />
(D)<br />
NaOH(aq)/Br 2<br />
OH<br />
Br<br />
Br<br />
O<br />
Br Br<br />
O<br />
Br<br />
the<br />
Bromination take place at ortho & para position<br />
due to activation of benzene ring by –OH group.<br />
10. The reagent(s) used for softening the temporary<br />
hardness of water is (are) -<br />
(A) Ca 3 (PO 4 ) 2 (B) Ca(OH) 2<br />
(C) Na 2 CO 3<br />
(D) NaOCl<br />
Ans. [B, C, D]<br />
11. Aqueous solutions of HNO 3 , KOH, CH 3 COOH<br />
and CH 3 COONa of identical concentrations are<br />
provided. The pair(s) of solutions which from a<br />
buffer upon mixing is(are)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 67<br />
MAY 2010
(A) HNO 3 and CH 3 COOH<br />
(B) KOH and CH 3 COONa<br />
(C) HNO 3 and CH 3 COONa<br />
(D) CH 3 COOH and CH 3 COONa<br />
Ans. [C, D]<br />
Sol. Acidic buffer is made up of weak acid & it's<br />
conjugate ion<br />
12. In the Newman projection for 2,2-dimethylbutane<br />
X<br />
Ans. [B, D]<br />
H 3 C<br />
CH 3<br />
H H<br />
Y<br />
X and Y can respectively be<br />
(A) H and H (B) H and C 2 H 5<br />
(C) C 2 H 5 and H (D) CH 3 and CH 3<br />
CH 3<br />
Sol. CH 3 –C–CH 2 –CH 3<br />
1 2 3 4<br />
CH 3<br />
along C 2 –C 3 Bond<br />
X = CH 3<br />
Y = CH 3<br />
CH 3<br />
CH 3 –C–CH 2 –CH 3<br />
1 2 3 4<br />
CH 3<br />
along C 1 –C 2<br />
X = H<br />
Y = C 2 H 5<br />
≡<br />
≡<br />
H<br />
C H 3<br />
H<br />
C H 3<br />
CH 3<br />
CH 3<br />
H<br />
C 2 H 5<br />
CH 3<br />
H<br />
Option [D]<br />
CH 3<br />
H<br />
Option [B]<br />
13. Among the following, the intensive property is<br />
(properties are)<br />
(A) molar conductivity<br />
(B) electromotive force<br />
(C) resistance<br />
(D) heat capacity<br />
Ans. [A, B]<br />
SECTION – III<br />
Paragraph Type<br />
This section contains 2 paragraphs. Based upon the<br />
first paragraph 2 multiple choice questions and based<br />
upon the second paragraph 3 multiple choice questions<br />
have to be answered. Each of these questions has four<br />
choices (A), (B), (C) and (D) out of which ONLY ONE<br />
is correct.<br />
Paragraph for Questions No. 14 to 15<br />
The concentration of potassium ions inside a<br />
biological cell is at least twenty times higher than<br />
the outside. The resulting potential difference<br />
across the cell is important in several processes<br />
such as transmission of nerve impulses and<br />
maintaining the ion balance. A simple model for<br />
such a concentration cell involving a metal M is :<br />
M(s) | M + (aq; 0.05 mloar) || M + (aq; 1 molar) |<br />
M(s)<br />
For the above electrolytic cell the magnitude of<br />
the potential |E cell | = 70 mV.<br />
14. For the above cell<br />
(A) E cell < 0; ∆G > 0 (B) E cell > 0; ∆G < 0<br />
(C) E cell < 0; ∆Gº > 0 (D) E cell > 0; ∆Gº < 0<br />
Ans. [B]<br />
Sol. For concentration cell<br />
k 0.05<br />
E cell = – log n 1<br />
[where k is +ve constant]<br />
= + ve<br />
∴ E cell > 0 ; ∆G < 0<br />
15. If the 0.05 molar solution of M + is replaced by a<br />
0.0025 molar M + solution, then the magnitude of<br />
the cell potential would be<br />
(A) 35 mV<br />
(B) 70 mV<br />
(C) 140 mV<br />
(D) 700 mV<br />
Ans. [C]<br />
Sol.<br />
k 0.0025<br />
E cell = – log n 1<br />
⎡ k 0.05⎤<br />
= 2 ⎢– log ⎥ = 2 × 70 = 140<br />
⎣ n 1 ⎦<br />
Paragraph for Question No. 16 to 18<br />
Copper is the most noble of the first row<br />
transition metals and occurs in small deposits in<br />
several countries. Ores of copper include<br />
chalcanthite (CuSO 4 . 5H 2 O), atacamite (Cu 2 Cl<br />
(OH) 3), cuprite (Cu 2 O), copper glance (Cu 2 S) and<br />
malachite (Cu 2 (OH) 2 CO 3 ). However, 18% of the<br />
world copper production come from the ore<br />
chalcopyrite (CuFeS 2 ). The extraction of copper<br />
from chalcopyrite involves partial roasting,<br />
removal of iron and self-reduction.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 68<br />
MAY 2010
16. Partial roasting of chalcopyrite produces<br />
(A) Cu 2 S and FeO (B) Cu 2 O and FeO<br />
(C) CuS and Fe 2 O 3 (D) Cu 2 O and Fe 2 O 3<br />
Ans. [A]<br />
17. Iron is removed from chalcopyrite as -<br />
(A) FeO<br />
(B) FeS<br />
(C) Fe 2 O 3 (D) FeSiO 3<br />
Ans. [D]<br />
18. In self-reduction, the reducing species is -<br />
(A) S (B) O 2–<br />
(C) S 2– (D) SO 2<br />
Ans. [C]<br />
Sol.<br />
1. O 3<br />
2. Zn, H 2O<br />
O<br />
O<br />
O<br />
(Only one product)<br />
OH –<br />
Heating<br />
23. Amongst the following, the total number of<br />
compound soluble in aqueous NaOH is<br />
O<br />
O<br />
O<br />
OH<br />
SECTION – IV<br />
Integer Type<br />
This section contains. TEN questions. The answer to<br />
each questions is a single digit integer ranging from 0<br />
to 9. The correct digit below the question number in the<br />
ORS is to be bubbled.<br />
235<br />
19. The number of neutrons emitted when 92 U<br />
undergoes controlled nuclear fission to<br />
142<br />
90<br />
54 Xeand 38Sr<br />
is<br />
Ans. [4]<br />
Sol.<br />
235<br />
1<br />
0<br />
142<br />
92 U + n → 54 Xe + 38Sr<br />
+ x n +<br />
235 + 1 = 142 + 90 + x<br />
x = 4<br />
90<br />
1<br />
0<br />
1<br />
1<br />
y p<br />
H 3 C<br />
N CH 3<br />
NO 2<br />
Ans. [4]<br />
Sol.<br />
COOH OCH 2 CH 3<br />
CH 2 OH<br />
OH<br />
N<br />
H 3 C CH 3<br />
COOH<br />
CH 2 CH 3<br />
CH 2 CH 3<br />
OH<br />
OH<br />
COOH<br />
20. The total number of basic groups in the following<br />
form of lysine is<br />
⊕<br />
H 3 N–CH 2 –CH 2 –CH 2 –CH 2 O<br />
Ans. [2]<br />
H 2 N<br />
CH–C<br />
21. The total number of cyclic isomers possible for a<br />
hydrocarbon with the molecular formula C 4 H 6 is<br />
Ans. [5]<br />
Sol.<br />
, , , ,<br />
22. In the scheme given below, the total number of<br />
intramolecular aldol condensation products form<br />
'Y' is<br />
Ans. [1]<br />
1. O 3 1. NaOH(aq)<br />
Y<br />
2. Zn, H 2O 2. heat<br />
O<br />
OH<br />
N<br />
H 3 C CH 3<br />
,<br />
,<br />
COOH<br />
24. Amongst the following, the total number of<br />
compounds whose aqueous solution turns red<br />
litmus paper blue is<br />
KCN K 2 SO 4 (NH 4 ) 2 C 2 O 4 NaCl Zn(NO 3 ) 2<br />
FeCl 3 K 2 CO 3 NH 4 NO 3 LiCN<br />
Ans. [3]<br />
Sol. Basic salt are KCN, LiCN, K 2 CO 3<br />
25. Based on VSEPR theory, the number of 90 degree<br />
F–Br–F angles in BrF 5 is<br />
Ans. [0]<br />
26. The value of n in the molecular formula<br />
Be n Al 2 Si 6 O 18 is<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 69<br />
MAY 2010
Ans. [3]<br />
Sol. Be 3 Al 2 Si 6 O 18<br />
Si 6 O 18 –12 ion<br />
27. A student performs a titration with different<br />
burettes and finds titre values of 25.2 mL, 25.25<br />
mL and 25.0 mL. The number of significant<br />
figures in the average titre value is<br />
Ans. [3]<br />
Sol.<br />
25 .2 + 25.25 + 25.0<br />
= 25.15<br />
3<br />
Significant figure = 3<br />
Significant figure in the answer can not be more<br />
than least significant figure any given value.<br />
28. The concentration of R in the reaction R → P was<br />
measured as a function of time and the following<br />
data is obtained :<br />
[R] 1.0 0.75 0.40 0.10<br />
(molar)<br />
t (min.) 0.0 0.05 0.12 0.18<br />
The order of the reaction is<br />
Ans. [0]<br />
Sol.<br />
R → P<br />
Assume zero order<br />
R = [R 0 ] – kt<br />
[ R 0 ] −[R]<br />
k =<br />
t<br />
1− 0.75<br />
k 1 = = 5<br />
0.5<br />
1− 0.4<br />
k 2 = = 5<br />
0.12<br />
1− 0.1<br />
k 3 = = 5<br />
0.18<br />
∴ order of reaction should be zero.<br />
MATHEMATICS<br />
SECTION – I<br />
Single Correct Choice Type<br />
This section contains 8 multiple choice questions. Each<br />
question has four choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
29. Let f , g and h be real-valued functions defined on<br />
2<br />
x x<br />
the interval [0, 1] by f(x) =<br />
2 −<br />
e + e ,<br />
2<br />
x x<br />
g(x) =<br />
2<br />
2<br />
−<br />
2 x x<br />
xe + e and h(x) =<br />
2 −<br />
x e + e .<br />
If a , b and c denote, respectively, the absolute<br />
maximum of f, g and h on [0, 1], then<br />
Ans.[D]<br />
(A) a = b and c ≠ b<br />
(C) a ≠ b and c ≠ b<br />
Sol. f '(x) = 2x( ⎜<br />
⎛ x 2 ⎟<br />
⎞<br />
⎝<br />
− − x<br />
e e<br />
2<br />
⎠<br />
x 2<br />
g'(x) = e ( 2x 2x 1)<br />
2 − +<br />
h'(x) =<br />
3 2<br />
2 x e x<br />
(B) a = c and a ≠ b<br />
(D) a = b = c<br />
Q all f'(x) , g'(x), h'(x) are positive so all attains<br />
absolute maxima at x = 1<br />
So Q f (1) = g(1) = h(1) = e + e –1 = a = b = c<br />
30. Let p and q be real numbers such that p ≠ 0, p 3 ≠<br />
q and p 3 ≠ – q. If α and β are non zero complex<br />
numbers satisfying α + β = – p and α 3 + β 3 = q,<br />
Ans.[B]<br />
then a quadratic equation having β<br />
α and α<br />
β as its<br />
roots is -<br />
(A) (p 3 + q) x 2 – (p 3 + 2q) x + (p 3 + q) =0<br />
(B) (p 3 + q) x 2 – (p 3 – 2q) x + (p 3 + q) =0<br />
(C) (p 3 – q) x 2 – (5p 3 – 2q) x + (p 3 – q) =0<br />
(D) (p 3 – q) x 2 – (5p 3 + 2q) x + (p 3 – q) =0<br />
Sol. α + β = – p ……….(1)<br />
α 3 + β 3 = q<br />
⇒ (α+ β) ( α 2 + β 2 – αβ) = q<br />
⇒ (α+ β) (( α + β) 2 – 3αβ) = q<br />
⇒ (–p) (p 2 – 3αβ) = q<br />
Now S =<br />
αβ =<br />
α β<br />
+<br />
β α<br />
(Sum of root) S =<br />
q + p<br />
3p<br />
=<br />
p<br />
3<br />
p<br />
…..(2)<br />
2<br />
( α + β)<br />
− 2αβ<br />
αβ<br />
3<br />
3<br />
− 2q<br />
+ q<br />
using (1) and (2)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 70<br />
MAY 2010
(Product of root) P =<br />
α β . = 1<br />
β α<br />
31. Equation of the plane containing the straight line<br />
x y z<br />
= = and perpendicular to the plane<br />
2 3 4<br />
x y z<br />
containing the straight lines = = and<br />
3 4 2<br />
x y z<br />
= = is<br />
4 2 3<br />
(A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0<br />
Ans.[C]<br />
Sol.<br />
(C) x – 2y + z = 0 (D) 5x + 2y – 4z = 0<br />
Plane passing through origin (0, 0, 0) and normal<br />
vector to plane is perpendicular to 3 î + 4ĵ + 2kˆ<br />
,<br />
4 î + 2ĵ + 3kˆ and 2 î + 3ĵ + 4kˆ<br />
i.e. normal vector<br />
to plane is î − 2ĵ + kˆ so equation to plane is x –<br />
2y + z = 0.<br />
32. If the angle A, B and C of a triangle are in an<br />
arithmetic progression and if a, b and c denote the<br />
lengths of the sides opposite to A, B and C<br />
respectively, then the value of the expression<br />
Ans.[D]<br />
Sol.<br />
a<br />
sin 2C<br />
c<br />
(A) 2<br />
1<br />
c<br />
+ sin 2A is -<br />
a<br />
(B)<br />
3<br />
2<br />
(C) 1 (D) 3<br />
a c<br />
sin 2C + sin 2A<br />
c a<br />
a<br />
c<br />
=<br />
2sin C cosC +<br />
c<br />
a<br />
a cosC + ccos A<br />
R<br />
2sin AcosA<br />
b 2R sin B<br />
= = R R<br />
= 3<br />
33. Let ω be a complex cube root of unity with ω<br />
≠ 1. A fair die is thrown three times If r 1 , r 2 and r 3<br />
are the numbers obtained on the die, then the<br />
r1 r2<br />
r3<br />
probability that ω + ω + ω = 0 is<br />
Sol. n(s) = 6 3<br />
Similarly total number of elements in events set is<br />
48<br />
48 12 2<br />
= = = 216 54 9<br />
34. Let P, Q, R and S be the points on the plane with<br />
position vectors – 2î<br />
− ĵ, 4 î , 3 î + 3ĵ<br />
and<br />
Ans.[A]<br />
Sol.<br />
r 1 r 2 r 3<br />
1 1<br />
2<br />
3<br />
4<br />
5<br />
6<br />
x<br />
3,6<br />
2,5<br />
x<br />
3,6<br />
2,5<br />
− 3 î + 2ĵ respectively. The quadrilateral PQRS<br />
must be a -<br />
(A) Parallelogram, which is neither a rhombus<br />
nor a rectangle<br />
(B) Square<br />
(C) Rectangle, but not a square<br />
(D) Rhombus, but not a square<br />
PQ = 6i +<br />
RS = 6i +<br />
2 1<br />
2<br />
3<br />
4<br />
5<br />
6<br />
3 1<br />
2<br />
3<br />
4<br />
5<br />
6<br />
j<br />
j<br />
RQ = i − 3j<br />
3,6<br />
x<br />
1,4<br />
3,6<br />
x<br />
1,4<br />
2,5<br />
1,4<br />
x<br />
2,5<br />
1,4<br />
x<br />
(A) 18<br />
1<br />
(B) 9<br />
1<br />
(C) 9<br />
2<br />
(D) 36<br />
1<br />
SP = i − 3j<br />
Ans.[C]<br />
PQ ≠ RQ (∴ not a rhombus or a rectangle)<br />
PQ || RS<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 71<br />
MAY 2010
RQ || SP<br />
Also PQ . RQ ≠ 0<br />
∴ PQRS is not a square<br />
⇒ PQRS is a parallelogram<br />
35. The value of<br />
Ans.[B]<br />
Sol.<br />
x lim →0<br />
t ln(1<br />
+ t)<br />
dt<br />
3 ∫<br />
is<br />
4<br />
t + 4<br />
x1 x 0<br />
(A) 0 (B) 12<br />
1<br />
1<br />
(C) 24<br />
Use L'hospital rule in<br />
1 t ln(1<br />
x +<br />
Lim<br />
x → 0 3 ∫ 0 4<br />
x t + 4<br />
x ln(1<br />
+ x)<br />
= Lim<br />
x → 0 4 2<br />
+<br />
= Lim<br />
=<br />
(x<br />
4)3x<br />
ln(1<br />
+ x)<br />
x<br />
x → 0 .3(x<br />
2 + 4 )<br />
1 1 =<br />
3.4 12<br />
t)<br />
dt<br />
1<br />
(D) 64<br />
36. The number of 3 × 3 matrices A whose entries are<br />
either 0 or 1 and for which the system<br />
⎡x⎤<br />
⎡1⎤<br />
A<br />
⎢ ⎥<br />
=<br />
⎢ ⎥<br />
⎢<br />
y<br />
⎥ ⎢<br />
0<br />
⎥<br />
has exactly two distinct solution, is -<br />
⎢⎣<br />
z⎥⎦<br />
⎢⎣<br />
0⎥⎦<br />
(A) 0 (B) 2 9 – 1<br />
(C) 168 (D) 2<br />
Ans.[A]<br />
SECTION – II<br />
Multiple Correct Choice Type<br />
This section contains 5 multiple choice questions. Each<br />
question has four choices (A), (B), (C) and (D), out of<br />
which ONE OR MORE may be correct.<br />
37. Let ABC be a triangle such that ∠ ACB = 6<br />
π and<br />
let a, b and c denote the lengths of the sides<br />
opposite to A, B and C respectively. The value(s)<br />
of x for which a = x 2 + x +1, b = x 2 – 1 and c = 2x<br />
+ 1 is (are)<br />
(A) − ( 2 + 3)<br />
(B) 1+ 3<br />
(C) 2 + 3<br />
(D) 4 3<br />
Ans.[B]<br />
Sol.<br />
As sum of two sides is always greater than third<br />
side, So x > 1<br />
Now<br />
2 +<br />
π a b – c<br />
Cos = 6 2ab<br />
(a – b) 2ab – c<br />
=<br />
23 2 +<br />
2ab<br />
2 2<br />
(a – b) – c<br />
3 – 2 =<br />
ab<br />
3 – 2 =<br />
(x<br />
2<br />
(x + 2)<br />
2<br />
2<br />
2<br />
+ x + 1)<br />
2<br />
– (2x + 1)<br />
(x<br />
2<br />
2<br />
–1)<br />
–3<br />
3 – 2 =<br />
2<br />
x + x + 1<br />
x 2 + x + 1 =<br />
3<br />
2 – 3<br />
x 2 + x + 1 = 3(2 + 3 )<br />
x 2 + x – 5 – 3 3 = 0<br />
(x – (1 + 3 ) (x + (2 + 3 )) = 0<br />
x = 1 + 3 , – (2 + 3 )<br />
So x = 3 + 1<br />
38. Let A and B be two distinct points on the<br />
parabola y 2 = 4x. If the axis of the parabola<br />
touches a circle of radius r having AB as its<br />
diameter, then the slope of the line joining A and<br />
B can be -<br />
1<br />
1<br />
(A) − (B)<br />
r<br />
r<br />
2<br />
2<br />
(C) (D) −<br />
r r<br />
Ans.[C, D]<br />
2<br />
2<br />
Sol. Let A ( t 1 ,2t1)<br />
B ( t 2,2t<br />
2 )<br />
2(t<br />
Slope =<br />
2 − t1)<br />
2<br />
=<br />
2 2<br />
t 2 − t1<br />
t 1 + t 2<br />
Equation of circle will be<br />
2 2<br />
(x − t1<br />
) (x − t 2)<br />
+ ( y − 2t1<br />
) ( y − 2t<br />
2)<br />
= 0<br />
2 2 2 2<br />
x + y – x(t1 + t 2)<br />
– 2y(t1 + t 2)<br />
+ 2 2<br />
t 1 t 2 + 4t 1 t 2<br />
= 0<br />
As it touches x axis so<br />
2 2 2<br />
2 2 (t t )<br />
t 1 t 2 + 4t 1 t 2 =<br />
1 + 2<br />
4<br />
2 2<br />
4 4 2 2<br />
4t 1 t 2 + 16 t 1 t 2 = t 1 + t 2 + 2 t 1 t 2<br />
2 2<br />
( ) 2<br />
t1 − t 2 = 16 t 1 t 2 . . . (1)<br />
AB is diameter so<br />
2 2 2<br />
1 t 2)<br />
( t − + 4 (t 1 – t 2 ) 2 = 4r 2 . . . (2)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 72<br />
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From (1) and (2)<br />
4 t 1 t 2 + (t 1 – t 2 ) 2 = r 2<br />
(t 1 + t 2 ) 2 = r 2<br />
t 1 + t 2 = ± r<br />
2<br />
∴ Slope = ± r<br />
x (1 − x)<br />
39. The value(s) of<br />
∫ 2<br />
1+<br />
x<br />
(A)<br />
(C) 0<br />
22<br />
− π<br />
7<br />
Ans.[A]<br />
1 4<br />
x (1 − x)<br />
Sol. I =<br />
∫ 2<br />
1+<br />
x<br />
0<br />
4<br />
1<br />
0<br />
4<br />
4<br />
dx is (are) -<br />
2<br />
(B) 105<br />
71 3π<br />
(D) −<br />
15 2<br />
1 8 7 6 5 4<br />
x − 4x + 6x − 4x + x<br />
=<br />
∫<br />
dx<br />
2<br />
0 1+<br />
x<br />
1<br />
6 5 4 2 4<br />
=<br />
∫<br />
( x − 4x + 5x − 4x + 4 − )dx<br />
2<br />
0<br />
1+<br />
x<br />
22<br />
= − π<br />
7<br />
40. Let z 1 and z 2 be two distinct complex numbers let<br />
z = (1–t) z 1 + tz 2 for some real number t with 0 < t<br />
< 1.<br />
If Arg (w) denotes the principal argument of a<br />
nonzero complex number w, then<br />
(A) | z – z 1 | + | z – z 2 | = | z 1 – z 2 |<br />
(B) Arg (z – z 1 ) = Arg (z – z 2 )<br />
(C)<br />
z − z1<br />
z2<br />
− z1<br />
z − z1<br />
z2<br />
− z1<br />
= 0<br />
(D) Arg (z – z 1 ) = Arg (z 2 – z 1 )<br />
Ans.[A,C,D]<br />
Sol. t =<br />
z − z<br />
z −<br />
2<br />
1<br />
z1<br />
z − z1<br />
So, = t e io V t ∈(0, 1)<br />
z2<br />
− z1<br />
So Geometrically<br />
z 1 z z 2<br />
So option A, C, D are true.<br />
41. Let f be a real valued function defined on the<br />
x<br />
interval (0, ∞ ) by f(x) = ln<br />
x +<br />
∫<br />
1 + sin t dt .<br />
0<br />
Then which of the following statement(s) is (are)<br />
true ?<br />
(A) f"(x) exists for all x ∈ (0, ∞ )<br />
(B) f '(x) exists for all x ∈ (0, ∞ ) and f ' is<br />
continuous on (0, ∞ ), but not differentiable<br />
on (0, ∞ )<br />
(C) there exists α > 1 such that |f '(x)| < | f (x) |<br />
for all x ∈ (α, ∞ )<br />
(D) there exists β > 0 such that |f '(x)| + | f '(x) | ≤<br />
β for all x ∈ (0, ∞ )<br />
Sol.[B,C]<br />
x<br />
f(x) = ln x +<br />
∫<br />
1+<br />
sin t d t<br />
0<br />
1<br />
1 x x<br />
f ′(x) = + 1+<br />
sin x = + |cos + sin |<br />
x<br />
x 2 2<br />
(A is not correct)<br />
1 x<br />
+ 1+ sin x < ln x + 1 sin t dt<br />
x<br />
∫<br />
+<br />
0<br />
Q ln x > x<br />
1 for some x = α ∀ α > 1<br />
and<br />
x<br />
1+ sin x <<br />
∫<br />
1 + sin t dt<br />
for some x = α ∀ α > 1<br />
so option C is correct<br />
0<br />
SECTION – III<br />
Paragraph Type<br />
This section contains 2 paragraphs. Based upon the<br />
first paragraph 2 multiple choice questions and based<br />
upon the second paragraph 3 multiple choice questions<br />
have to be answered. Each of these questions has four<br />
choices (A), (B), (C) and (D) out of which ONLY ONE<br />
is correct.<br />
Paragraph for Question No. 42 to 43<br />
The circle x 2 + y 2 x 2<br />
– 8x = 0 and hyperbola – 9<br />
y 2<br />
= 1 intersect at the points A and B<br />
4<br />
42. Equation of a common tangent with positive slope<br />
to the circle as well as to the hyperbola is -<br />
(A) 2x – 5 y – 20 = 0<br />
(B) 2x – 5 y + 4 = 0<br />
(C) 3x – 4y + 8 = 0<br />
(D) 4x – 3y + 4 = 0<br />
Ans.[B]<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 73<br />
MAY 2010
Sol. y = m(x – 4) ± 4<br />
y = mx ± 9m<br />
2 − 4<br />
– 4m ± 4<br />
2<br />
m<br />
2<br />
1+ m<br />
1+ = ± 9m<br />
2 − 4<br />
16m 2 + 16 + 16m 2 m 32 m<br />
2<br />
2<br />
1+ m = 9m 2 – 4<br />
m 32m 1+ m = – 23m 2 – 20<br />
1024m 2 + 1024 m 4 = 529m 4 + 400 + 920 m 2<br />
495 m 4 + 104 m 2 – 400 = 0<br />
(5m 2 – 4) (99m 2 + 100) = 0<br />
∴ m 2 4<br />
2<br />
= ∴ m = ± 5 5<br />
So tangent with positive slope<br />
y =<br />
2 4 x ±<br />
5 5<br />
2x – 5 y ± 4 = 0<br />
43. Equation of the circle with AB as its diameter is<br />
(A) x 2 + y 2 – 12 x + 24 = 0<br />
(B) x 2 + y 2 + 12 x + 24 = 0<br />
(C) x 2 + y 2 + 24 x – 12 = 0<br />
(D) x 2 + y 2 – 24x – 12 = 0<br />
Ans.[A]<br />
Sol. x 2 + y 2 – 8x = 0<br />
4x 2 – 9y 2 = 36<br />
⎛ ⎞<br />
x 2 + ⎜<br />
4x<br />
2 − 36<br />
⎟ – 8x = 0<br />
⎝ 9 ⎠<br />
13x 2 – 72x – 36 = 0<br />
(x – 6) (13x + 6) = 0<br />
−6<br />
x = 6, 13<br />
x = 6, y = ± 12<br />
∴ Equation of required circle<br />
(x – 6) 2 + (y – 12 ) (y + 12 ) = 0<br />
x 2 + y 2 – 12x + 24 = 0<br />
Paragraph for Question No. 44 to 46<br />
Let p be an odd prime number and T p be the<br />
following set of 2 × 2 matrices :<br />
⎪⎧<br />
⎡a<br />
b⎤<br />
⎪⎫<br />
= ⎨A<br />
= ⎢ ⎥ : a, b,c ∈ { 0,1,2,......, p −1} ⎬<br />
⎪⎩ ⎣c<br />
a⎦<br />
⎪⎭<br />
T p<br />
44. The number of A in T p such that A is either<br />
symmetric or skew-symmetric or both, and det<br />
(A) divisible by p is -<br />
(A) (p – 1) 2 (B) 2(p – 1)<br />
(C) (p –1) 2 + 1 (D) 2p –1<br />
Ans.[D]<br />
Sol.<br />
|A | = a 2 – bc<br />
if A is symmetric b = c<br />
then | A | = (a + b) (a – b)<br />
So a & b can attain 2(p – 1) solution<br />
It A is skew symmetric then a = b = c = 0<br />
So total no. of solution = 2p – 2 + 1 = 2p – 1<br />
45. The number of A in T p such that the trace of A is<br />
not divisible by p but det (A) is divisible by p is<br />
[Note : The trace of a matrix is the sum of its<br />
diagonal entries.]<br />
(A) (p – 1) (p 2 – p + 1) (B) p 3 – (p – 1) 2<br />
(C) (p –1) 2 (D) (p –1) (p 2 – 2)<br />
Ans.[C]<br />
46. The number of A in T p such that det (A) is not<br />
divisible by p is -<br />
(A) 2p 2<br />
(B) p 3 – 5p<br />
(C) p 3 –3p (D) p 3 – p 2<br />
Ans. [D]<br />
SECTION – IV<br />
Integer type<br />
This section contains TEN paragraphs. The answer to<br />
each question is a single-digit integer, ranging from 0 to<br />
9. The correct digit below the question number in the<br />
ORS is to be bubbled.<br />
47. Let f be a real-valued differentiable function on R<br />
(the set of all real numbers) such that f(1) = 1. If<br />
the y-intercept of the tangent at any point P(x, y)<br />
on the curve y = f(x) is equal to the cube of the<br />
abscissa of P, then the value of f(–3) is equal to –<br />
Ans.[9]<br />
dy<br />
Sol. Y – y = ( X – x) dx<br />
dy<br />
y – x = x<br />
3<br />
dx<br />
dy<br />
x – y = – x<br />
3<br />
dx<br />
T y – = – x<br />
2<br />
T x<br />
1<br />
2<br />
I.F. = x<br />
1<br />
y =<br />
x ∫ − xdx + C<br />
y − x 2<br />
= + C<br />
x 2<br />
1<br />
3<br />
1 = − + C ∴ C =<br />
2<br />
2<br />
∴ y =<br />
x( −x<br />
2<br />
2 +<br />
3)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 74<br />
MAY 2010
2 +<br />
x( −x<br />
3)<br />
∴ f(x) =<br />
2<br />
−3 ( −9<br />
+ 3)<br />
f(–3) =<br />
= 9<br />
2<br />
48. The number of values of θ in the interval<br />
⎛ π π ⎞<br />
nπ<br />
⎜−<br />
, ⎟ such that θ ≠ for n = 0 , ± 1, ± 2<br />
⎝ 2 2 ⎠<br />
5<br />
and tan θ = cot 5 θ as well as sin 2 θ = cos 4 θ is –<br />
Ans.[3]<br />
π<br />
Sol. tan θ = cot 5θ ⇒ θ = (2n + 1) 12<br />
π π 5π<br />
So θ = ± , ± , ± 12 4 12<br />
1<br />
∴ Sin 2θ = cos 4θ ⇒ sin 2θ = 2<br />
or – 1<br />
π 5π –π<br />
only θ = , , satisfies the given<br />
12 12 4<br />
conditions<br />
So total number of solution = 3<br />
49. The maximum value of the expression<br />
1<br />
is<br />
2<br />
2<br />
sin θ + 3sin θcosθ + 5cos θ<br />
Sol.[2] f (θ) =<br />
sin<br />
2<br />
1<br />
θ + 3sin θcos<br />
θ + 5cos<br />
1<br />
=<br />
2<br />
1+ 4cos θ + 3sin θcosθ<br />
1<br />
=<br />
3<br />
1+ 2(1 + cos 2θ)<br />
+ sin 2θ<br />
2<br />
1<br />
=<br />
3<br />
3 + 2cos 2θ + sin 2θ<br />
2<br />
1<br />
So f(θ) max = = 2<br />
9<br />
3 – 4 +<br />
4<br />
50. If a r and b r are vector is space given by<br />
r î − 2ĵ 2î + ĵ+<br />
3kˆ<br />
a = and bˆ = , then the value of<br />
5<br />
14<br />
r r r r r r<br />
2a + b . a × b × a − 2b is<br />
( ) [( ) ( )]<br />
Ans.[5]<br />
Sol. | a | = | b | = 1& a ⋅ b = 0<br />
( 2a + b) ⋅[(a<br />
× b) × (a – 2b)]<br />
2 2<br />
= ( 2 a + b) ⋅[b<br />
+ 2a] = | b | + 4 | a | = 5<br />
2<br />
θ<br />
51. The line 2x + y = 1 is tangent to the hyperbola<br />
2<br />
2<br />
x y − =1. If this line passes through the point<br />
2 2<br />
a b<br />
of intersection of the nearest directrix and the x-<br />
axis, then the eccentricity of the hyperbola is<br />
Ans.[2]<br />
Sol. 1 = 4a 2 – b 2 ... (1)<br />
2a<br />
= 1<br />
e<br />
e<br />
a = ... (2)<br />
2<br />
also b 2 = a 2 (e 2 – 1) ... (3)<br />
(1) & (3)<br />
1 = 4a 2 – a 2 e 2 + a 2 ⇒ 1 = 5a 2 – a 2 e 2<br />
2<br />
5 e e<br />
⇒ 1 = –<br />
4 4<br />
⇒ e 4 – 5e 2 + 4 = 0<br />
⇒ (e 2 – 4) (e 2 – 1) = 0<br />
∴ e = 2<br />
4<br />
52. If the distance between the plane Ax – 2y + z = d<br />
and the plane containing the lines<br />
x −1<br />
y − 2 z − 3 x − 2 y − 3 z − 4<br />
= = and = = is<br />
2 3 4 3 4 5<br />
Ans.[6]<br />
Sol.<br />
î<br />
2<br />
3<br />
6 , then | d | is –<br />
ĵ<br />
3<br />
4<br />
kˆ<br />
4<br />
5<br />
= î(<br />
− 1) − ĵ( −2)<br />
+ kˆ( −1)<br />
Plane is normal to vector î − 2ĵ+<br />
kˆ<br />
1(X – 1) – 2 (Y –2) + 1(Z – 3) = 0<br />
X – 2Y + Z = 0<br />
| d |<br />
6 = ⇒ | d | = 6<br />
6<br />
53. For any real number x, let [x] denote the largest<br />
integer less than or equal to x. Let f be a real<br />
valued function defined on the interval [–10, 10]<br />
by<br />
Ans.[4]<br />
f (x)<br />
⎧ x −[x]<br />
= ⎨<br />
⎩1<br />
+ [x] − x<br />
Then the value of<br />
if [x]<br />
is odd<br />
if [x]is even<br />
2<br />
π 10<br />
∫ f (x)<br />
10 − 10<br />
⎧ {x} V[x]is odd<br />
Sol. f(x) = ⎨<br />
⎩1<br />
= {x} V[x]is even<br />
graph of y = f(x) is<br />
cos πx dx is<br />
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56. The number of all possible values of θ, where<br />
0 < θ < π, for which the system of equations<br />
–5 –4 –3 –2 –1 0 1 2 3<br />
(y + z) cos 3 θ = (xyz) sin 3θ<br />
Q f(x) & cos πx both are even functions<br />
2cos3θ<br />
2sin 3θ<br />
x sin 3θ = +<br />
2 10<br />
y z<br />
π<br />
So, I = 10 ∫<br />
f (x) cos πx dx<br />
(xyz) sin3θ = (y + 2z) cos 3θ + y sin 3θ)<br />
−10<br />
have a solution (x<br />
2 10<br />
0 , y 0 , z 0 ) with y 0 z 0 ≠ 0, is<br />
π<br />
=<br />
5 ∫<br />
f (x)cos( π x)dx<br />
Ans.[3]<br />
0<br />
Sol. (xyz) sin 3θ + y (– cos 3θ) + z (– cos 3θ) = 0<br />
∴ f(x) & cos πx both are periodic then<br />
(xyz) sin 3θ + y (–2 sin 3θ) + z (–2 cos 3θ) = 0<br />
2<br />
(xyz) sin 3θ + y (– cos 3θ − sin 3θ ) + z (– 2cos<br />
I = π 2 ∫<br />
f (x) cos(πx) dx<br />
3θ) = 0<br />
0<br />
For y<br />
⎡<br />
1<br />
2<br />
0 z 0 ≠ 0 ⇒ Nontrivial solution<br />
⎤<br />
= π<br />
2<br />
⎢ − π + − π ⎥<br />
⎢∫(<br />
1 x)cos x( x)dx<br />
∫(x<br />
1)cos( x)dx<br />
sin 3θ<br />
− cos3θ<br />
− cos3θ<br />
⎥<br />
⎣ 0<br />
1<br />
⎦<br />
sin 3θ<br />
− 2sin 3θ<br />
− 2cos3θ<br />
= 0<br />
2 ⎡2<br />
+ 2⎤<br />
sin 3θ<br />
− cos3θ − sin 3θ<br />
− 2cos3θ<br />
= π ⎢ 2 ⎥ = 4<br />
⎣ π ⎦<br />
1 cos3θ<br />
1<br />
2 π 2<br />
+ isin<br />
sin 3θ cos3θ 1 2sin 3θ<br />
2 = 0<br />
.<br />
3 3<br />
1 cos3θ + sin 3θ<br />
2<br />
Then the number of distinct complex number z<br />
sin3θ cos3θ [(4sin 3θ – 2 cos 3θ – 2sin 3θ) –<br />
2<br />
z + 1 ω ω<br />
(2cos 3θ – cos 3θ – sin 3θ) + 2 cos 3θ – 2 sin 3θ]<br />
2<br />
satisfying ω z + ω 1 = 0 is equal to =<br />
= 0<br />
2<br />
ω 1 z + ω<br />
⇒ (sin3θ cos3θ) [2 sin3θ – 2cos 3θ – cos 3θ + sin<br />
3θ + 2cos 3θ – 2sin 3θ] = 0<br />
On solving the determinant<br />
It become z 3 = 0<br />
⇒ (sin3θ cos3θ) (sin3θ – cos 3θ) = 0<br />
So no. of solutions = 1<br />
π 2π<br />
⇒ sin 3θ = 0 ⇒ θ = , 3 3<br />
These two donot<br />
satisfy system<br />
infinite geometric series whose first term is k<br />
π π 5π 7π of equations<br />
k −1<br />
1 ⇒ cos 3θ = 0 ⇒ θ = , , , and the common ratio is . Then the value<br />
6 2 6 6<br />
k!<br />
k<br />
100 2 100<br />
2<br />
of + ∑ ( k − 3k + 1S ) k<br />
100!<br />
k=<br />
1<br />
is –<br />
π 5π 9π<br />
⇒ sin 3θ = cos 3θ ⇒ 3θ = , , 4 4 4<br />
K<br />
π 5π 3π<br />
⇒ θ = , , = 3<br />
12 12 4<br />
K<br />
No. of solutions = 3<br />
100<br />
∑ |(k 2 – 3k + 1)S k |<br />
K=<br />
1<br />
PHYSICS<br />
100 ( k<br />
– 3k + 1)<br />
= 1 + 1 + ∑<br />
k –1<br />
K=<br />
3<br />
SECTION – I<br />
= 2 + ∑<br />
k – 1 k<br />
Single Correct Choice Type<br />
–<br />
k – 2 k – 1<br />
54. Let ω be the complex number cos<br />
Ans.[1]<br />
Sol.<br />
55. Let S k , k = 1, 2, ……, 100, denote the sum of the<br />
Ans.[4]<br />
Sol. S k =<br />
= 2 + 2 –<br />
99<br />
This section contains 8 multiple choice questions. Each<br />
question has 4 choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 76<br />
MAY 2010
57. A thin uniform annular disc (see figure) of mass<br />
M has outer radius 4R and inner radius 3R. The<br />
work required to take a unit mass from point P on<br />
its axis to infinity is-<br />
3R<br />
P<br />
2GM<br />
2GM<br />
(A) (4 2 − 5)<br />
(B) − (4 2 − 5)<br />
7R<br />
7R<br />
GM<br />
2GM<br />
(C)<br />
(D) ( 2 − 1)<br />
4R<br />
5R<br />
Ans. [A]<br />
Sol. ∆W ext = U 2 – U 1<br />
for unit +ve mass<br />
U 1 = V 1 and U 2 = V 2 = 0<br />
V 1 =<br />
∫ dV = ∫ − Gdm<br />
2 2 1/<br />
(r + 16R )<br />
2<br />
=<br />
∫ − GM2πrdr<br />
2 2 2 1/<br />
7πR<br />
(r + 16R )<br />
2<br />
Put r 2 + 16R 2 = t 2<br />
4 2R<br />
2GM 2GM<br />
V 1 = −<br />
2 ∫<br />
dt = − (4 2 − 5)<br />
7R<br />
7R<br />
5R<br />
2GM<br />
∆W ext = U 2 – U 1 = (4 2 − 5)<br />
7R<br />
58. A block of mass m is on an inclined plane of<br />
angle θ. The coefficient of friction between the<br />
block and the plane is µ and tan θ > µ. The block<br />
is held stationary by applying a force P parallel to<br />
the plane. The direction of force pointing up the<br />
plane is taken to be positive. As P is varied from<br />
P 1 = mg(sinθ – µ cosθ) to P 2 = mg(sinθ + µ cosθ),<br />
the frictional force f versus P graph will look<br />
like –<br />
f<br />
P<br />
θ<br />
(A) P 2<br />
P 1<br />
(C)<br />
f<br />
P 1<br />
P 2<br />
P<br />
P<br />
4R<br />
4R<br />
(B)<br />
f<br />
f<br />
P 1<br />
P 2<br />
(D) P 1 P 2<br />
P<br />
P<br />
Ans.<br />
Sol.<br />
[A]<br />
In the given range block is in equilibrium so<br />
P – mg sin θ + f = 0<br />
f = mg sinθ – P<br />
Equation of straight line with negative slope.<br />
59. A real gas behaves like an ideal gas if its -<br />
(A) pressure and temperature are both high<br />
(B) pressure and temperature are both low<br />
(C) pressure is high and temperature is low<br />
(D) pressure is low and temperature is high<br />
Ans.<br />
Sol.<br />
[D]<br />
Reason : PV = nRT holds true in case of low<br />
pressure and high temperature conditions.<br />
60. Consider a thin square sheet of side L and<br />
thickness t, made of a material of resistivity ρ.<br />
The resistance between two opposite faces, shown<br />
by the shaded areas in the figure is-<br />
Ans.<br />
Sol.<br />
t<br />
(A) directly proportional to L<br />
(B) directly proportional to t<br />
(C) independent of L<br />
(D) independent of t<br />
[C]<br />
ρL<br />
R = A<br />
R = t<br />
ρ<br />
ρL<br />
= Lt<br />
L<br />
R is independent of L<br />
61. Incandescent bulbs are designed by keeping in<br />
mind that the resistance of their filament increases<br />
with the increase in temperature. If at room<br />
temperature, 100 W, 60 W and 40 W bulbs have<br />
filament resistances R 100 , R 60 and R 40 ,<br />
respectively, the relation between these<br />
resistances is-<br />
1 1 1<br />
(A) = +<br />
R R<br />
(B) R 100 = R 40 + R 60<br />
Ans.<br />
100 40<br />
R<br />
60<br />
(C) R 100 > R 60 > R 40<br />
[D]<br />
V 2<br />
Sol. Rated power =<br />
R<br />
1<br />
R ∝<br />
Rated power<br />
∴<br />
1<br />
R<br />
1<br />
R<br />
><br />
1 2<br />
R<br />
3<br />
><br />
1<br />
R<br />
1<br />
R<br />
><br />
><br />
1<br />
1<br />
100 60<br />
R<br />
40<br />
(D)<br />
1<br />
R<br />
><br />
P 1 > P 2 > P 3<br />
1<br />
R<br />
+<br />
1<br />
100 60<br />
R<br />
40<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 77<br />
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62. To verify Ohm's law, student is provided with a<br />
test resistor R T , a high resistance R 1 , a small<br />
resistance R 2 , two identical galvanometers G 1 and<br />
G 2 , and a variable voltage source V. The correct<br />
to carry out the experiment is-<br />
Ans.<br />
(A)<br />
(B)<br />
(C)<br />
(D)<br />
[C]<br />
G 1<br />
G 1<br />
G 1<br />
R 2<br />
R T<br />
G 2<br />
R 1<br />
V<br />
G 1<br />
R 1<br />
R T<br />
G 2<br />
R 2<br />
V<br />
V<br />
V<br />
R 1<br />
R T<br />
G 2<br />
R 2<br />
Sol. Converted ammeter =<br />
Converted ammeter =<br />
R 2<br />
R T<br />
G 2<br />
Voltmeter should be connected in parallel to R T<br />
and Ammeter should be connected in series with<br />
R T .<br />
G<br />
R 1<br />
R 2<br />
G<br />
R 1<br />
63. A thin flexible wire of length L is connected to<br />
two adjacent fixed points and carries a current i in<br />
the clockwise direction, as shown in the figure.<br />
When the system is put in a uniform magnetic<br />
field of strength B going into plane of the paper,<br />
the wire takes the shape of a circle. The tension<br />
in the wire isx<br />
x x x x x x<br />
x x x x x x x<br />
x x x x x x x<br />
x x x x x x x<br />
x x x x x x x<br />
Ans.<br />
Sol.<br />
(A) IBL<br />
(C)<br />
[C]<br />
T<br />
IBL<br />
2π<br />
dθ<br />
T cos<br />
2<br />
i<br />
dθ<br />
dθ<br />
2T sin<br />
2<br />
T = tension<br />
dθ<br />
iB × R dθ = 2Tsin<br />
2<br />
T = iBR<br />
2πR = L<br />
L<br />
R =<br />
∴ T =<br />
2π<br />
x x x x x x x<br />
(B)<br />
(D)<br />
dθ<br />
T cos<br />
2<br />
IBL<br />
π<br />
IBL<br />
4π<br />
F magnetic = iB×Rdθ<br />
iBL<br />
2π<br />
64. An AC voltage source of variable angular<br />
frequency ω and fixed amplitude V 0 is connected<br />
in series with a capacitance C and an electric bulb<br />
of resistance R (inductance zero). When ω is<br />
increased -<br />
(A) the bulb glows dimmer<br />
(B) the bulb glows brighter<br />
(C) total impedance of the circuit is unchanged<br />
(D) total impedance of the circuit increases<br />
Ans. [B]<br />
R C<br />
Sol.<br />
V 0 , ω<br />
~<br />
T<br />
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2 ⎛ 1 ⎞<br />
Z = R + ⎜ ⎟<br />
⎝ ωC<br />
⎠<br />
ω increased z decreased<br />
∴ current in circuit increase<br />
∴ Bulb glow brighter.<br />
2<br />
SECTION – II<br />
Multiple Correct Choice Type<br />
This section contains 5 multiple correct answer(s) type<br />
questions. Each question has 4 choices (A), (B), (C) and<br />
(D), out of which ONE OR MORE is/are correct.<br />
65. A student uses a simple pendulum 1 m length to<br />
determine g, the acceleration due to gravity. He<br />
uses a stop watch the least count of 1 sec for this<br />
records 40 seconds for 20 oscillations. For this<br />
observation, which of the following statement(s)<br />
is (are) true ?<br />
(A) Error ∆T in measuring T, the time period, is<br />
0.05 seconds<br />
(B) Error ∆T in measuring T, the time period, is 1<br />
second<br />
(C) Percentage error in the determination of g is<br />
5%<br />
(D) Percentage error in the determination of g is<br />
2.5 %<br />
Ans. [A,C]<br />
Total time(t)<br />
Sol. Time period (T) =<br />
no. of oscillations<br />
So<br />
∆ T ∆ 1sec<br />
= =<br />
T tt<br />
40sec<br />
1<br />
∆T = × 2 = 0.05 sec<br />
40<br />
l<br />
T = 2π ; T 2 =<br />
g<br />
4π<br />
2 l<br />
g<br />
∆ g ∆ = + 2<br />
g ll ∆ ;<br />
TT<br />
∆T = 0.05 sec ;<br />
∆ l 2× 0.05<br />
Putting we get =<br />
l 2<br />
2<br />
4π<br />
l<br />
; g =<br />
2<br />
T<br />
T = 2 sec.<br />
∆ g × 100 = 5 %<br />
g<br />
∆ l = 0<br />
l<br />
= 0.05<br />
66. A few electric field lines for a system of two<br />
charges Q 1 and Q 2 fixed at two different points on<br />
the x-axis are shown in the figure. These lines<br />
suggest that<br />
;<br />
Ans.<br />
Sol.<br />
Q<br />
(A) |Q 1 | > |Q 2 |<br />
(B) |Q 1 | < |Q 2 |<br />
(C) at a finite distance to the left of Q 1 the electric<br />
field is zero<br />
(D) at a finite distance to the right of Q 2 the<br />
electric field is zero<br />
[A,D]<br />
Number of field lines emitting from Q 1 is more<br />
than number of field lines reaching at Q 2<br />
So | Q 1 | > | Q 2 |<br />
and if so E r at a point which is right to Q 2 will be<br />
zero.<br />
67. One mole of an ideal gas in initial state A<br />
undergoes a cyclic process ABCA, As shown in<br />
figure. Its pressure at A is P 0 . Choose the correct<br />
option(s) from the following<br />
V<br />
4V<br />
B<br />
Ans.<br />
Sol.<br />
V<br />
C A<br />
T<br />
T<br />
(A) Internal energies at A and B are the same<br />
(B) Work done by the gas in process AB is P 0 V 0<br />
ln 4<br />
(C) Pressure at C is P 0 /4<br />
T<br />
(D) Temperature at C is 0<br />
4<br />
[A,B]<br />
From figure<br />
AB → isothermal process<br />
So T A = T B ⇒ Internal energies will be same.<br />
⎛ V ⎞<br />
2<br />
W AB = nRT 0 ln<br />
⎜<br />
⎟ = P0 V 0 ln 4<br />
⎝ V1<br />
⎠<br />
It is not given that line BC passes through origin.<br />
So we can't find pressure or temperature at point C.<br />
68. A <strong>Point</strong> mass 1 kg collides elastically with a<br />
stationary point mass of 5 kg. After their<br />
collision, the 1 kg mass reverses its direction and<br />
moves with a speed of 2 ms –1 . Which of the<br />
following statement(s) is (are) correct for the<br />
system of these two masses ?<br />
(A) Total momentum of the system is 3 kg ms –1<br />
(B) Momentum of 5 kg mass after collision is 4<br />
kg ms –1<br />
(C) Kinetic energy of the centre of mass is 0.75 J<br />
(D) Total kinetic energy of the system is 4 J<br />
Q<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 79<br />
MAY 2010
Ans.<br />
Sol.<br />
[A,C]<br />
1kg<br />
v 1<br />
5kg<br />
Before collision<br />
1kg 5kg<br />
2m/sec<br />
v2<br />
After collision<br />
Collision is elastic so<br />
v 2 + 2 = v 1<br />
......(i)<br />
Conservation of momentum,<br />
1 × v 1 + 0 = – 2 × 1 + 5 × v 2<br />
....(ii)<br />
Solving<br />
v 1 = 3 m/sec<br />
v 2 = 1 m/sec<br />
Total momentum p r<br />
system<br />
= 1 × v 1 = 3 kg-m/sec<br />
Momentum of 5 kg = 5 × v 2 = 5 kg-m/sec<br />
1 × 3 + 5×<br />
0<br />
v CM =<br />
= 0.5 m/sec<br />
6<br />
1<br />
2<br />
k CM = × (M1 + M<br />
2<br />
) vCM<br />
= 0.75 joule<br />
2<br />
1<br />
k system = × 1 × 9 = 4.5 joule.<br />
2<br />
69. A rat OP of monochromatic light is incident on<br />
the face AB of prism ABCD near vertex B at an<br />
incident angle of 60º (see figure). If the refractive<br />
index of the material of the prism is 3 , which<br />
of the following is (are) correct ?<br />
B<br />
Ans.<br />
60º<br />
60º<br />
135º<br />
90º 75º<br />
A<br />
D<br />
(A) The ray gets totally internally reflected at face<br />
CD<br />
(B) The ray comes out through face AD<br />
(C) The angle between the incident ray and the<br />
emergent ray is 90º<br />
(D) The angle between the incident ray and the<br />
emergent ray is 120º<br />
[A,B,C]<br />
Sol. Refraction at first surface AB :<br />
C<br />
sin 60º<br />
sin r<br />
= 1<br />
3<br />
O<br />
60º<br />
B<br />
A<br />
60º<br />
90º<br />
C<br />
135º<br />
30º = r<br />
60º<br />
r 1<br />
E<br />
r 2<br />
45º<br />
it hits at E<br />
By geometry angle r 1 = 45º, r 2 = 45º<br />
1 1<br />
We know sinθ C = and sin 45º =<br />
3<br />
2<br />
1 1 ><br />
2 3<br />
so 45º > θ C<br />
So total internal reflection occurs.<br />
After reflection angle of incidence at AD will be<br />
30º so ray comes out making an angle 60º with<br />
the normal at AD.<br />
Final Ray<br />
60º<br />
60º30º<br />
Incident Ray<br />
SECTION – III<br />
Paragraph Type<br />
This section contains 2 paragraphs.. Based upon the<br />
first paragraph 2 multiple choice question and based<br />
upon the second paragraph 3 multiple choice question<br />
have to be answered. Each of these questions has four<br />
choices (A), (B), (C) and (D) for its answer, out of<br />
which ONLY ONE is correct.<br />
Paragraph for Question No. 70 to 71<br />
Electrical resistance of certain materials, knows<br />
as superconductors, changes abruptly from a<br />
nonzero value to zero as their temperature is<br />
lowered below a critical temperature T C (0). An<br />
interesting property of superconductors is that<br />
their critical temperature becomes smaller than T C<br />
(0) if they are placed in a magnetic field, i.e. the<br />
critical temperature T C (B) is a function of the<br />
magnetic field strength B. The dependence of T C<br />
(B) on B is shown in the figure.<br />
T C (B)<br />
T C (0)<br />
D<br />
r = 30º<br />
O<br />
B<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 80<br />
MAY 2010
70. In the graphs below, the resistance R of a<br />
superconductor is shown as a function of its<br />
temperature T for two different magnetic fields B 1<br />
(solid line) and B 2 (dashed line). If B 2 is larger<br />
than B 1 , which of the following graphs shows the<br />
correct variation of R with T in these fields ?<br />
(A)<br />
R<br />
R<br />
O<br />
B 2<br />
B 2<br />
B 1<br />
T<br />
Paragraph for Question No. 72 to 74<br />
When a particle of mass m moves on the x-axis in<br />
a potential of the form V(x) = kx 2 , it performs<br />
simple harmonic motion. The corresponding time<br />
m<br />
period is proportional to , as can be seen<br />
k<br />
easily using dimensional analysis. However, the<br />
motion of a particle can be periodic even when<br />
its potential energy increases on both sides of x =<br />
0 in a way different from kx 2 and its total energy<br />
is such that the particle does not escape to<br />
infinity. Consider a particle of mass m moving<br />
on the x-axis. Its potential energy is V(x) = αx 4<br />
(α > 0) for | x | near the origin and becomes a<br />
constant equal to V 0 for | x | ≥ X 0 (see figure)<br />
V(x)<br />
(B)<br />
(C)<br />
(D)<br />
R<br />
R<br />
O<br />
O<br />
O<br />
B 1<br />
B 1<br />
Ans. [A]<br />
Sol. B 2 > B 1<br />
so T C (B 2 ) < T C (B 1 )<br />
Dashed Solid<br />
line line<br />
Resistance ∝ Temperature above critical .<br />
B 1<br />
B 2<br />
B 2<br />
71. A superconductor has T C (0) = 100 K. When a<br />
magnetic field of 7.5. Tesla is applied, its T C<br />
decreases to 75 K. For this material one can<br />
definitely say that when<br />
(A) B = 5 Tesla, T C (B) = 80 K<br />
(B) B = 5 Tesla, 75 K < T C (B) < 100 K<br />
(C) B = 10 Tesla, 75 K < T C (B) < 100 K<br />
(D) B = Tesla, T C (B) = 70 K<br />
Ans. [B]<br />
Sol. T C (0) = 100 K, B = 0<br />
B = 7.5 tesla<br />
T C (B) = 75<br />
If B = 5 Tesla, T C (B) should be greater than 75 K<br />
T<br />
T<br />
T<br />
72. If the total energy of the particle is E, it will<br />
perform periodic motion only if-<br />
(A) E < 0 (B) E > 0<br />
(C) V 0 > E > 0 (D) E > V 0<br />
Ans.<br />
Sol.<br />
[C]<br />
Energy Total should be less than maximum<br />
potential energy<br />
so E < V 0 and E > 0.<br />
73. For periodic motion of small amplitude A, the<br />
time period T of this particle is proportional to-<br />
(A)<br />
(C)<br />
Ans [B]<br />
Sol. V(x) = α x 4<br />
[V(x)]<br />
[α] =<br />
4<br />
[x]<br />
X 0<br />
V 0<br />
m<br />
A (B)<br />
α<br />
A α (D)<br />
m<br />
=<br />
[ML T<br />
4<br />
[L ]<br />
2 −2<br />
]<br />
1<br />
A<br />
x<br />
m<br />
α<br />
1 α<br />
A m<br />
= [ML –2 T –2 ]<br />
Time period ∝ (Amplitude) x (α) y (Mass) z<br />
[T] = [L] x [ML –2 T –2 ] y [M] 2<br />
1 1<br />
Solving x = –1, y = − , z =<br />
2 2<br />
T = A –1 α –1/2 M 1/2 =<br />
1<br />
A<br />
M<br />
α<br />
74. The acceleration of this particle for | x | > X 0 is -<br />
(A) proportional to V 0<br />
V0<br />
(B) proportional to<br />
mX<br />
(C) proportional to<br />
0<br />
V0<br />
mX<br />
0<br />
(D) zero<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 81<br />
MAY 2010
Ans. [D]<br />
Sol. for |x| > x 0<br />
U = constant<br />
dU<br />
F = – = 0<br />
dx<br />
acceleration = zero.<br />
SECTION – IV<br />
Integer Type<br />
This section contains TEN questions. The answer to<br />
each question is a single-digit integer, ranging from 0 to<br />
9. The correct digit below the question number in the<br />
ORS is to be bubbled.<br />
75. A stationary source is emitting sound at a fixed<br />
frequency f 0 , which is reflected by two cars<br />
approaching the source. The difference between<br />
the frequencies of sound reflected from the cars is<br />
1.2% of f 0 . What is the difference in the speeds<br />
of the car in km per hour) to the nearest integer ?<br />
The cars are moving at constant speeds much<br />
smaller than the speed of sound which is 330 m/s.<br />
Ans. [7]<br />
v 1 v 2<br />
Sol.<br />
v<br />
2<br />
+ v(v<br />
f 0<br />
*<br />
S<br />
2<br />
⎛ v + v1<br />
⎞<br />
f 1 = f 0 ⎜ ⎟<br />
⎝ v ⎠<br />
⎛ v ⎞ f<br />
f 2 = f 1<br />
⎜<br />
⎟<br />
0(v<br />
+ v1)<br />
=<br />
⎝ v − v 1 ⎠ (v − v1)<br />
f0(v<br />
+ v2)<br />
f 2'<br />
=<br />
(v − v )<br />
f '<br />
2<br />
−f<br />
f<br />
0<br />
2<br />
2<br />
(v + v2<br />
) (v + v1)<br />
= – = 0.012<br />
(v − v ) (v − v )<br />
(v + v2)(v<br />
− v1)<br />
− (v + v1)(v<br />
− v2)<br />
= 0.012<br />
(v − v )(v − v )<br />
− v1)<br />
− v1v2<br />
− v + v1v<br />
(v − v )(v − v )<br />
2<br />
2<br />
2<br />
1<br />
2<br />
2<br />
1<br />
− v(v1<br />
− v2)<br />
= 0.012<br />
2v(v2<br />
− v1)<br />
(v − v )(v − v<br />
1<br />
2<br />
1<br />
)<br />
= 0.012<br />
2<br />
(v2 v1)<br />
v<br />
− = 0.012<br />
18<br />
(v 2 – v 1 ) = 0.006 × 330 × 5<br />
= 7.128<br />
76. The focal length of thin biconvex lens is 20 cm.<br />
When an object is moved from a distance of 25<br />
cm in front of it to 50 cm, the magnification of its<br />
m<br />
25<br />
image changes from m 25 to m 50 . The ratio<br />
m50<br />
is ?<br />
Ans. [6]<br />
Sol. m =<br />
f<br />
f + u<br />
20<br />
m 25 = = –4 ; m 50 =<br />
20 − 25<br />
m 25 = 6<br />
m<br />
50<br />
20 2<br />
= –<br />
20 − 50 3<br />
77. An α-particle and a proton are accelerated from<br />
rest by a potential difference of 100V. After this,<br />
their de-Broglie wavelengths are λ α and λ p<br />
respectively. The ratio<br />
λ p<br />
λ α<br />
, to the nearest integer<br />
is ?<br />
Ans. [3]<br />
Sol. After accelerating through V 0 KE of a particle<br />
becomes = qV 0 evolts<br />
so<br />
KE α = 200 eV<br />
KE p = 100 eV<br />
h<br />
λ debroglied =<br />
2MKE<br />
λ P<br />
=<br />
λ α<br />
MαKE<br />
M<br />
α<br />
p KE p<br />
=<br />
= 2 × 1.414 ~ = 3<br />
4×<br />
200<br />
1×<br />
100<br />
= 2 2<br />
78. When two identical batteries of internal resistance<br />
1Ω each are connected in series across a resistor<br />
R, the rate of heat produced in R is J 1 . When the<br />
same batteries are connected in parallel across R,<br />
the rate is J 2 . If J 1 = 2.25 J 2 then the value of R is<br />
Ω is ?<br />
Ans. [4]<br />
Sol.<br />
R<br />
i 1<br />
1Ω<br />
R<br />
1Ω<br />
E E<br />
2E<br />
i 1 =<br />
R + 2<br />
2<br />
⎛ 2E ⎞<br />
J 1 = ⎜ ⎟ . R<br />
⎝ R + 2 ⎠<br />
i 2<br />
E<br />
E eq = E<br />
E<br />
r eq = 0.5Ω<br />
1Ω<br />
1Ω<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 82<br />
MAY 2010
i 2 =<br />
From J 1 = 2.25 J 2<br />
2<br />
2<br />
⎛ 2E ⎞ ⎛ E ⎞<br />
⎜ R<br />
R 2<br />
⎟ = 2.25 R<br />
⎝ +<br />
⎜<br />
⎠<br />
R 0.5<br />
⎟<br />
⎝ + ⎠<br />
2<br />
R + 2<br />
= 1.5<br />
R + 0. 5<br />
2R + 1 = 1.5 R + 3<br />
0.5 R = 2<br />
R =<br />
2 = 4 Ω<br />
0.5<br />
E<br />
R + 0.5<br />
⎛ E ⎞<br />
J 2 =<br />
⎜ R<br />
R 0.5<br />
⎟<br />
⎝ + ⎠<br />
79. Two spherical bodies A(radius 6 cm) and B<br />
(radius 18 cm) are at temperature T 1 and T 2 ,<br />
respectively. The maximum intensity in the<br />
emission spectrum of A is at 500 nm and in that<br />
of B is at 1500 nm. Considering them to be black<br />
bodies, what will be the ratio of the rate of total<br />
energy radiated by A to that of B ?<br />
Ans. [9]<br />
Sol. λ A = 500 nm<br />
λ B = 1500 nm<br />
λ A T A = λ B T B<br />
T =<br />
T T<br />
1<br />
2<br />
T A = 3 … (i)<br />
B<br />
rA<br />
1<br />
=<br />
rB<br />
3<br />
From Stefan's law<br />
4<br />
E σ(4πr<br />
2 )(T )<br />
=<br />
A 1 ⎛ 1 ⎞<br />
= ⎜<br />
4 ⎟⎠ × (3) 4 = 9<br />
E σ(4gr<br />
2 )(T ) ⎝ 3<br />
A<br />
B<br />
B<br />
2<br />
80. When two progressive waves y 1 = 4sin(2x – 6t)<br />
⎛ π ⎞<br />
and y 2 = 3 sin ⎜2 x − 6t − ⎟ are superimposed,<br />
⎝ 2 ⎠<br />
the amplitude of the resultant wave is ?<br />
Ans. [5]<br />
Sol. y 1 = 4 sin (2x –6t)<br />
y 2 = 3 sin (2x –6t –π/2)<br />
π<br />
φ = 2<br />
2<br />
2<br />
A res =<br />
2 2<br />
A1<br />
2 1 2<br />
2<br />
+ A + 2A A cosφ<br />
A res = 3 + 4 + 0 = 5<br />
2<br />
81. A 0.1 kg mass is suspended from a wire of<br />
negligible mass. The length of the wire is 1m and<br />
its cross-sectional area is 4.9 × 10 –7 m 2 . If the<br />
mass is pulled a little in the vertically downward<br />
direction and released, it performs simple<br />
harmonic motion of angular frequency 140 rad/s.<br />
If the Young's modulus of the material of the wire<br />
is n × 10 9 Nm –2 , the value of n is ?<br />
Ans. [4]<br />
Sol.<br />
x<br />
strain = l<br />
x<br />
stress = Y<br />
strain<br />
stress = Yx (l = 1 m)<br />
F = Yx<br />
A<br />
F = Ayx<br />
Ayx = ma<br />
a =<br />
140 =<br />
AYx<br />
m<br />
140 = 70 n<br />
n = 4<br />
; ω =<br />
–7<br />
4.9×<br />
10 × n × 10<br />
0.1<br />
AY<br />
m<br />
82. A binary star consists of two stars (mass 2.2 M S )<br />
and B(mass 11 M S ), where M S is the mass of the<br />
sun. They are separated by distance d and are<br />
rotating about their centre of mass, which is<br />
stationary. The ratio of the total angular<br />
momentum of the binary star to the angular<br />
momentum of star B about the centre of mass is ?<br />
Ans. [6]<br />
Sol.<br />
r A<br />
d<br />
r B<br />
9<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 83<br />
MAY 2010
2<br />
L A = m A ω r A<br />
2<br />
L B = m B ωr B<br />
Ratio (K) =<br />
r<br />
r<br />
m<br />
=<br />
m<br />
A<br />
L A + L<br />
L<br />
11<br />
=<br />
2.2<br />
B<br />
A B<br />
=<br />
B<br />
5<br />
B<br />
=<br />
Ratio (K) = 5<br />
1 × 5 2 + 1 = 6<br />
2<br />
L A m + 1 =<br />
ArA<br />
2<br />
LB<br />
mBrB<br />
+ 1<br />
Ans. [8]<br />
Sol. The amount of heat required to raise the temp<br />
from –5°C to 0°C.<br />
Q 1 = m × 2100 × 10 –3 × 5 = 10.5 m Joule<br />
The amount of heat required to melt 1 gm<br />
ice = 10 –3 × 3.36 × 10 5 = 336 J<br />
420 = 336 + 10.5 m<br />
10.5 = 84<br />
m = 8 gm.00<br />
83. Gravitational acceleration on the surface of a<br />
planet is 11<br />
6 g, where g is the gravitational<br />
acceleration on the surface of the earth. The<br />
2<br />
average mass density of the planet is times 3<br />
that of the earth. If the escape speed on the<br />
surface of the earth is taken to be 11 km/s, the<br />
escape speed on the surface of the planet in km/s<br />
will be ?<br />
Ans. [3]<br />
Sol.<br />
υ p 2g pR<br />
p =<br />
υc<br />
2geR<br />
e<br />
(1)<br />
⇒<br />
M<br />
4<br />
πR<br />
3<br />
p<br />
3<br />
p<br />
= 3<br />
2<br />
=<br />
g p R p × …<br />
g<br />
e<br />
Me<br />
4<br />
πR<br />
3<br />
3<br />
e<br />
R<br />
e<br />
M P 2 R<br />
⇒ =<br />
M 3 R<br />
GM p 6 GM M<br />
2 =<br />
e p 6 R<br />
R p 11<br />
2 ⇒ =<br />
R M<br />
e<br />
c 11 R<br />
from (2) and (3)<br />
3<br />
p<br />
2 R 6<br />
3 =<br />
3 R e 11 R<br />
from (1) and (4)<br />
2<br />
p<br />
2<br />
e<br />
R R p 3 ⇒ =<br />
R e 226<br />
υ p 6 3 18 3<br />
= × = =<br />
υe<br />
11 226<br />
242 11<br />
v p = 3 km /sec.<br />
e<br />
2<br />
p<br />
2<br />
e<br />
3<br />
p<br />
3<br />
e<br />
… (4)<br />
… (2)<br />
…(3)<br />
84. A piece of ice (heat capacity = 2100 Jkg –1 ºC –1 and<br />
latent heat = 3.36 × 10 8 J kg –1 ) of mass m grams<br />
is at –5ºC at atmospheric pressure. It is given 420<br />
J of heat so that the ice starts melting. Finally<br />
when the ice-water mixture is in equilibrium, it is<br />
found that 1 gm of ice has melted. Assuming<br />
there is no other heat exchange in the process, the<br />
value of m is ?<br />
ATTITUDE<br />
• Great effort springs naturally from a great<br />
attitude.<br />
• Like success, failure is many things to many<br />
people. With Positive Mental Attitude, failure<br />
is a learning experience, a rung on the ladder,<br />
a plateau at which to get your thoughts in<br />
order and prepare to try again.<br />
• Your attitude, not your aptitude, will<br />
determine your altitude.<br />
• Develop an attitude of gratitude, and give<br />
thanks for everything that happens to you,<br />
knowing that every step forward is a step<br />
toward achieving something bigger and better<br />
than your current situation.<br />
• You can adopt the attitude there is nothing<br />
you can do, or you can see the challenge as<br />
your call to action.<br />
• "An optimist is a person who sees a green<br />
light everywhere, while the pessimist sees<br />
only the red stoplight... The truly wise person<br />
is colorblind."<br />
• Positive thinking will let you do everything<br />
better than negative thinking will.<br />
• You cannot control what happens to you, but<br />
you can control your attitude toward what<br />
happens to you, and in that, you will be<br />
mastering change rather than allowing it to<br />
master you.<br />
• You can do it if you believe you can!<br />
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<strong>IIT</strong>-<strong>JEE</strong> 2010<br />
PAPER-II (PAPER & SOLUTION)<br />
Time : 3 Hours Total Marks : 237<br />
Instructions :<br />
• The question paper consists of 3 Parts (Chemistry, Mathematics and Physics). And each part consists of four<br />
Sections.<br />
• For each question in Section I: you will be awarded 5 marks if you have darkened only the bubble corresponding<br />
to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus two (–2) mark will be<br />
awarded.<br />
• For each question in Section II: you will be awarded 3 marks if you darken the bubble corresponding to the<br />
correct answer and zero mark if no bubbles is darkened. No negative marks will be awarded for incorrect answers<br />
in this section.<br />
• For each question in Section III: you will be awarded 3 marks if you darken only the bubble corresponding to the<br />
correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be<br />
awarded.<br />
• For each question in Section IV: you will be awarded 2 marks for each row in which you have darkened the<br />
bubbles(s) corresponding to the correct answer. Thus, each question in this section carries a maximums of<br />
8 marks. There is no negative marks awarded for incorrect answer(s) in this section.<br />
CHEMISTRY<br />
SECTION – I<br />
Single Correct Choice Type<br />
This section contains 6 multiple choice questions. Each<br />
question has 4 choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
1. Assuming that Hund's rule is violated, the bond<br />
order and magnetic nature of the diatomic<br />
molecule B 2 is<br />
(A) 1 and diamagnetic<br />
(B) 0 and diamagnetic<br />
(C) 1 and paramagnetic<br />
(D) 0 and paramagnetic<br />
Ans. [A]<br />
Sol. 2<br />
1s<br />
*<br />
1s<br />
σ σ<br />
2<br />
σ<br />
2<br />
2 σ<br />
2s<br />
*<br />
2s<br />
2<br />
p<br />
π =<br />
2 x<br />
B. Ο = 2<br />
1 (6 – 4) = 1<br />
Diamagnetic<br />
2. The compounds P, Q and S<br />
COOH<br />
π 2p y<br />
OCH 3<br />
were separately subjected to nitration using<br />
HNO 3 /H 2 SO 4 mixture. The major product formed<br />
in each case respectively, is<br />
(A)<br />
(B)<br />
HO<br />
HO<br />
NO 2<br />
COOH<br />
O 2 N<br />
COOH<br />
NO 2<br />
H 3 C<br />
H 3 C<br />
NO 2<br />
O<br />
||<br />
C<br />
O<br />
NO 2<br />
OCH 3<br />
OCH 3<br />
HO<br />
P<br />
H 3 C<br />
Q<br />
O<br />
||<br />
C<br />
O<br />
O<br />
||<br />
C<br />
NO 2<br />
O<br />
S<br />
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(C)<br />
HO<br />
NO 2<br />
COOH<br />
O<br />
||<br />
C<br />
H 3 C<br />
O<br />
OCH 3<br />
NO 2<br />
NO 2<br />
(A) 39.27 % (B) 68.02%<br />
(C) 74.05% (D) 78.54%<br />
Ans. [D]<br />
Sol. Area of square = L 2<br />
(D)<br />
HO<br />
NO 2<br />
COOH<br />
H 3 C<br />
O<br />
||<br />
C<br />
O<br />
OCH 3<br />
NO 2<br />
NO 2<br />
& 4R =<br />
R =<br />
2 L<br />
L<br />
2 2<br />
area of circle<br />
% packing efficiency (η) = × 100<br />
area of square<br />
Ans.<br />
Sol.<br />
[C]<br />
COOH<br />
HNO 3 /H 2 SO 4<br />
COOH<br />
NO 2<br />
2× π R<br />
= × 100<br />
2<br />
L<br />
L<br />
2× π<br />
= 4×<br />
2 × 100<br />
2<br />
L<br />
2<br />
2<br />
= 78. 5%<br />
OH<br />
OH<br />
OCH 3 OCH 3<br />
NO<br />
HNO 3 /H 2 SO 2<br />
4<br />
CH 3 CH 3<br />
4. The species having pyramidal shape is<br />
(A) SO 3 (B) BrF 3<br />
Ans.<br />
(C)<br />
[D]<br />
2–<br />
SiO 3<br />
(D) OSF 2<br />
O<br />
||<br />
C HNO 3 /H 2 SO 4<br />
O<br />
3. The packing efficiency of the two-dimensional<br />
square unit cell shown below is<br />
O<br />
||<br />
C<br />
O<br />
NO 2<br />
Sol.<br />
F<br />
S<br />
F<br />
O<br />
5. In the reaction<br />
O (1) NaOH/Br 2<br />
H 3 C C O<br />
NH 2 (2) C<br />
Cl<br />
the structure of the product T is<br />
O<br />
(A) H 3 C C O<br />
O– C<br />
T<br />
L<br />
(B)<br />
NH<br />
C<br />
O<br />
CH 3<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 87<br />
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(C) H 3 C<br />
NH<br />
10 .5<br />
3 × N<br />
C<br />
108<br />
A<br />
Number of atoms of Ag in 1cc ⇒<br />
.5<br />
H 3 PO 4 H 2 SO 4 H 3 PO 3 H 2 CO 3 H 2 S 2 O 7<br />
× N A<br />
H<br />
108<br />
3 BO 3 H 3 PO 2 H 2 CrO 4 H 2 SO 3<br />
(D) H 3 C<br />
2/3<br />
O<br />
In 1cm<br />
O<br />
⎛10.5<br />
⎞<br />
, number of atoms of Ag = ⎜ × N A ⎟<br />
⎝ 108 ⎠<br />
C O<br />
In 10 –12 m 2 or 10 –8 cm 2 , number of atoms of<br />
NH–C<br />
2/3<br />
⎛10.5<br />
⎞<br />
Ag = ⎜ N A ⎟ × 10 –8<br />
⎝ 108 ⎠<br />
Ans. [C]<br />
2/3<br />
⎛<br />
23<br />
10.5 6.02 10 ⎞<br />
Sol.<br />
= ⎜<br />
× ×<br />
⎟ × 10 O<br />
108<br />
⎝<br />
⎠<br />
NaOH/Br 2<br />
CH 3 C CH NH<br />
Hoffmann's 3<br />
2<br />
= 1.5 × 10 7<br />
NH 2 degradation<br />
O<br />
||<br />
Thus, x = 7<br />
Cl O<br />
C<br />
NH–C<br />
8. Among the following, the number of elements<br />
CH 3 NH 2 +<br />
showing only one non-zero oxidation state is<br />
–HCl<br />
O, Cl, F, N. P, Sn, Tl, Na, TI<br />
CH 3<br />
Ans. [2]<br />
Sol. F & Na only show one non zero oxidation state<br />
that are – 1 & + 1 respectively.<br />
6. The complex showing a spin-only magnetic<br />
moment of 2.82 B.M. is<br />
9. One mole of an ideal gas is taken from a to b<br />
along two paths denoted by the solid and the<br />
(A) Ni(CO) 4 (B) [NiCl 4 ] 2–<br />
dashed lines as shown in the graph below. If the<br />
(C) Ni((PPh 3 ) 4 (D) [Ni(CN) 4 ] 2–<br />
work done along the solid line path is w s and that<br />
Ans. [B]<br />
along the dotted line paths is w d , then the integer<br />
Sol. [NiCl 4 ] –2 closest to the ratio w d \w s is<br />
Ni +2 1s 2 2s 2 2p 6 3s 2 3d 8 4s°<br />
4.<br />
4. a<br />
3.<br />
Cl – Cl – Cl – Cl –<br />
P 3.0<br />
(atm) 2.<br />
hyb = sp 3<br />
2.0<br />
1.5<br />
no. of unpaired electrons = 2<br />
1.0<br />
0.5<br />
b<br />
µ = 2 (4)<br />
= 8 = 2.82 BM<br />
0.0<br />
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0<br />
V(lit)<br />
Ans. [2]<br />
SECTION – II<br />
Sol. For solid line path show approxy isothermal<br />
Integer Type<br />
process<br />
This section contains a group of 5 questions. The<br />
5.5<br />
answer to each questions is a single digit integer<br />
∴ work done |W S | = 2.303 (PV) log<br />
ranging from 0 to 9. The correct digit below the<br />
.5<br />
question number in the ORS is to be bubbled.<br />
= 2.303 × 4 × .5 × log 11<br />
~– 4.79<br />
7. Silver (atomic weight = 108 g mol –1 ) has a<br />
for dashed line path work done<br />
density of 10.5 g cm –3 . The number of silver<br />
w<br />
atoms on a surface of area 10 –12 m 2 d = 4 × |2 – .5| + 1 × |3 – 2| + .5 × |5.5 –3|<br />
can be<br />
= 6 + 1 + 1.25 = 8.25<br />
expressed in scientific notation as y × 10 x . The<br />
w d 8.25<br />
value of x is<br />
∴ = = 1.71 ~– 2<br />
Ans. [7]<br />
w s 4. 8<br />
Sol.<br />
m 10. The total number of diprotic acids among the<br />
d = ⇒ 10.5 g/cc<br />
V following is<br />
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Ans. [6]<br />
Sol.<br />
O<br />
||<br />
HO – S – OH<br />
||<br />
O<br />
O O<br />
|| ||<br />
H–O – S – O– S – O – H<br />
||<br />
O<br />
||<br />
O<br />
O<br />
||<br />
HO – P – OH<br />
O<br />
OH<br />
Cr<br />
OH<br />
O<br />
OH<br />
O<br />
||<br />
S<br />
OH<br />
O<br />
||<br />
C<br />
OH<br />
11. Total number of geometrical isomers for the<br />
complex [RhCl(CO)(PPh 3 )(NH 3 )] is<br />
Ans. [3]<br />
Sol.<br />
Cl<br />
PPh 3<br />
Cl<br />
NH 3<br />
Rh<br />
Rh<br />
CO<br />
Cl<br />
NH 3<br />
NH 3<br />
Rh<br />
CO<br />
PPh 3<br />
CO<br />
PPh 3<br />
OH<br />
(B) H 3 C<br />
(C)<br />
H 3 C<br />
(D)<br />
H 3 C<br />
CH<br />
CH 3<br />
CH<br />
C<br />
O<br />
CH 3<br />
CH 2<br />
O<br />
CH<br />
C<br />
CH 3<br />
CH 2<br />
O<br />
Ans. [B]<br />
13. The compound R is<br />
(A) H 3 C<br />
H 3 C<br />
C<br />
CH 2<br />
O<br />
C<br />
OH<br />
C<br />
H<br />
H and H<br />
C<br />
O<br />
H H 3 C<br />
and<br />
H H<br />
and<br />
H<br />
C<br />
O<br />
C<br />
O<br />
H<br />
H<br />
SECTION – III<br />
Paragraph Type<br />
This section contains 2 paragraphs. Based upon each of<br />
the paragraph 3 multiple choice questions have to be<br />
answered. Each of these questions has four choices (A),<br />
(B), (C) and (D) out of which ONLY ONE is correct.<br />
Paragraph for questions No. 12 to 14<br />
(B) H 3 C<br />
H 3 C<br />
H 3 C<br />
C<br />
CH<br />
O<br />
C<br />
OH<br />
H<br />
Two aliphatic aldehydes P and Q react in the<br />
presence of aqueous K 2 CO 3 to give compound R,<br />
which upon treatment with HCN provides<br />
compound S. On acidification and heating, S<br />
gives the product shown below –<br />
H 3 C OH<br />
H 3 C<br />
12. The compounds P and Q respectively are -<br />
CH 3<br />
(A) H 3 C<br />
CH<br />
O<br />
C<br />
O<br />
O<br />
H and H 3 C<br />
C<br />
O<br />
H<br />
Ans.<br />
(C) H 3 C<br />
(D) H 3 C<br />
[A]<br />
CH 3<br />
CH<br />
CH<br />
CH 2<br />
CH 3<br />
CH<br />
H 3 C<br />
CH<br />
CH<br />
O<br />
C<br />
OH<br />
O<br />
C<br />
OH<br />
H<br />
H<br />
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14. The compound S is<br />
CH 3<br />
(A) H 3 C<br />
CH<br />
CH<br />
O<br />
C<br />
H<br />
CH 3<br />
(iv) CH 3 – C – CH 2<br />
OH<br />
H – C – CN<br />
OH<br />
⎯ ⎯ H +<br />
3 →<br />
O ⎯ ∆<br />
CH 3 – C — CH 2<br />
HC O<br />
OH C<br />
O<br />
Final product<br />
H 3 C<br />
(B)<br />
H3 C<br />
(C) H 3 C<br />
H 3 C<br />
(D)<br />
H3 C<br />
CH 3<br />
CH 2<br />
CN<br />
O<br />
C<br />
C<br />
H<br />
CH 2<br />
CN<br />
CH 3 CN<br />
CH<br />
CH<br />
CH<br />
CH 2<br />
OH<br />
CN<br />
C<br />
CH<br />
OH<br />
OH<br />
Paragraph for Questions No. 15 to 17<br />
The hydrogen like species Li 2+ is in a spherically<br />
symmetric state S 1 with one radial node. Upon<br />
absorbing light the ion undergoes transition to a<br />
state S 2 . The state S 2 has one radial node and its<br />
energy is equal to the ground state energy of the<br />
hydrogen atom.<br />
15. The state S 1 is –<br />
(A) 1s<br />
(C) 2p<br />
Ans. [B]<br />
Sol. Q One radial node<br />
∴ n – l – 1 = 1<br />
or n – l = 2<br />
l = 0<br />
n = 2<br />
Orbital name = 2s<br />
(B) 2s<br />
(D) 3s<br />
CH 2<br />
Ans. [D]<br />
Sol. (12 to 14)<br />
CH 3<br />
(i) CH 3 – CH<br />
C<br />
O<br />
H<br />
OH<br />
OH<br />
⎯ –<br />
⎯→ ⎯<br />
CH 3<br />
CH 3 – C Θ C<br />
O<br />
H<br />
16. Energy of the state S 1 in units of the hydrogen<br />
atom ground state energy is –<br />
(A) 0.75 (B) 1.50<br />
(C) 2.25 (D) 4.50<br />
Ans. [C]<br />
3<br />
Sol. S 1 = Energy of e of H in ground state ×<br />
2<br />
= 2.25 × energy of e of H in ground state<br />
2<br />
2<br />
(ii)<br />
CH 3<br />
CH 3<br />
H<br />
CH 3 – C Θ + C H CH 3 – C – CH 2<br />
C H<br />
O<br />
C OH<br />
H<br />
O<br />
O<br />
[R]<br />
CH 3<br />
CH 3<br />
(iii) CH 3 – C – CH 2 + HCN<br />
C OH<br />
H<br />
O<br />
C – CH 2<br />
CH 3 OH<br />
H – C – CN<br />
OH<br />
[S]<br />
17. The orbital angular momentum quantum number<br />
of the state S 2 is –<br />
(A) 0 (B) 1<br />
(C) 2 (D) 3<br />
Sol. [B]<br />
For S 2 = n – l – 1<br />
n – l = 2<br />
n = 3, l = 1<br />
Orbital = 3p ∴ l = 1<br />
2<br />
3<br />
[ S 2 = energy of e of H in ground state ×<br />
2 , n = 3]<br />
n<br />
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SECTION – IV<br />
Matrix Type<br />
This Section contains 2 questions. Each question has<br />
four statements (A, B, C and D) given in Column I and<br />
five statements (p, q, r, s and t) in Column II. Any<br />
given statement in Column I can have correct<br />
matching with one or more statement(s) given in<br />
Column II. For example, if for a given question,<br />
statement B matches with the statements given in q<br />
and r, then for that particular question, against<br />
statement B, darken the bubbles corresponding to q<br />
and r in the ORS.<br />
18. Match the reactions in Column I with appropriate<br />
options in Column II.<br />
Column I<br />
NaOH/<br />
H 2 O<br />
0°<br />
C<br />
(A) N 2 Cl + OH ⎯ ⎯⎯⎯⎯<br />
→<br />
N = N<br />
OH<br />
Ans. [A → r,s,t; B → t; C → p,q; D → r]<br />
19. All the compounds listed in Column I react with<br />
water. Match the result of the respective reactions<br />
with the appropriate options listed in Column II.<br />
Column I<br />
(A) (CH 3 ) 2 SiCl 2<br />
(B) XeF 4<br />
(C) Cl 2<br />
(D) VCl 5<br />
Column II<br />
(p) Hydrogen halide formation<br />
(q) Redox reaction<br />
(r) Reacts with glass<br />
(s) Polymerization<br />
(t) O 2 formation<br />
Ans. [A → p,s; B → p,q,r,t; C → p,q,t; D → p]<br />
MATHEMATICS<br />
SECTION – I<br />
Single Correct Choice Type<br />
This section contains 6 multiple choice questions. Each<br />
question has 4 choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
20. A signal which can be green or red with<br />
OH OH<br />
(B) H 3C – C – C – CH 3<br />
CH 3 CH 3<br />
H 2 SO<br />
⎯ 4<br />
⎯⎯→<br />
O<br />
probability 5<br />
4 and 5<br />
1 respectively, is received by<br />
station A and then transmitted to station B. The<br />
probability of each station receiving the signal<br />
correctly is 4<br />
3 . If the signal received at station B<br />
H 3 C<br />
C CH 3<br />
C<br />
CH3<br />
CH 3<br />
is green, then the probability that the original<br />
signal was green is<br />
3<br />
6<br />
(A) (B) 5 7<br />
(C)<br />
C<br />
O<br />
CH 3<br />
1. LiAlH<br />
2.H<br />
O<br />
4<br />
⎯⎯⎯ +<br />
→<br />
(D) HS Cl ⎯ Base ⎯⎯ →<br />
Column II<br />
(p) Racemic mixture<br />
(q) Addition reaction<br />
(r) Substitution reaction<br />
(s) Coupling reaction<br />
(t) Carbocation intermediate<br />
3<br />
S<br />
OH<br />
CH<br />
CH 3<br />
Ans.<br />
Sol.<br />
20<br />
(C) 23<br />
9<br />
(D) 20<br />
[C]<br />
Event (1) : original signal<br />
OG : Original signal is green<br />
OR : Original signal is red<br />
Event (2) : Signal received by A.<br />
AG : A received green<br />
AR : A received Red<br />
Event (3) : Signal received by B<br />
BG : B received green<br />
BR : B received Red<br />
⎛ BG ⎞<br />
P(OG).P⎜<br />
⎟<br />
⎛ OG ⎞<br />
P⎜<br />
⎟ =<br />
⎝ OG ⎠<br />
⎝ BG ⎠ ⎛ BG ⎞ ⎛ BG ⎞<br />
P(OG).P⎜<br />
⎟ + P(OR).P⎜<br />
⎟<br />
⎝ OG ⎠ ⎝ OR ⎠<br />
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=<br />
4 ⎡3<br />
⎢ .<br />
5 ⎣4<br />
3<br />
4<br />
4 ⎡3<br />
3 1 1 ⎤<br />
⎢ . + .<br />
5<br />
⎥<br />
⎣4<br />
4 4 4⎦<br />
1 1 ⎤ 1 ⎡1<br />
3<br />
+ . ⎥ + ⎢ . +<br />
4 4⎦<br />
5 ⎣ 4 4<br />
3<br />
.<br />
4<br />
1 ⎤<br />
4⎥<br />
⎦<br />
20<br />
= 23<br />
21. If the distance of the point P (1, –2, 1) from the<br />
plane x + 2y –2z = α, where α > 0, is 5, then the<br />
foot of the perpendicular from P to the plane is<br />
⎛ 8 4 7 ⎞<br />
⎛ 4 4 1 ⎞<br />
(A) ⎜ , ,− ⎟ (B) ⎜ , − , ⎟<br />
⎝ 3 3 3 ⎠ ⎝ 3 3 3 ⎠<br />
⎛ 1 2 10 ⎞<br />
⎛ 2 1 5 ⎞<br />
(C) ⎜ , , ⎟ (D) ⎜ , − , ⎟<br />
⎝ 3 3 3 ⎠ ⎝ 3 3 2 ⎠<br />
Ans. [A]<br />
x − 1 y + 2 z − |1−<br />
4 − 2 − α |<br />
Sol. = = = λ = 5<br />
1 2 − 21<br />
3<br />
foot (1 + λ, –2 + 2λ, 1 –2λ) |α + 5| = 15<br />
(1 + λ) + 2(–2 + 2λ) – 2 (1 –2λ) = 10<br />
α = 10 (correct), –20 (wrong)<br />
1 + λ – 4 + 4λ –2 + 4λ = 10<br />
9λ = 15 , ⇒ λ = 5/3<br />
⎛ 8 4 7 ⎞<br />
foot = ⎜ , ,− ⎟<br />
⎝ 3 3 3 ⎠<br />
22. Two adjacent sides of a parallelogram ABCD are<br />
given by AB= 2 î + 10 ĵ + 11 kˆ and AD = – î +<br />
2 ĵ + 2 kˆ The side AD is rotated by an acute angle<br />
Ans.<br />
Sol.<br />
α in the plane of the parallelogram so that AD<br />
becomes AD′. If AD′ makes a right angle with the<br />
side AB, then the cosine of the angle α is given<br />
by<br />
(A) 9<br />
8<br />
(C) 9<br />
1<br />
[B]<br />
D′<br />
– î + 2 ĵ + 2 kˆ<br />
A<br />
α<br />
D<br />
⎛ π ⎞<br />
⎜ − α⎟<br />
⎝ 2 ⎠<br />
(B)<br />
(D)<br />
2 î + 10 ĵ + 11 kˆ<br />
17<br />
9<br />
4 5<br />
9<br />
B<br />
C<br />
23. Let S = {1, 2, 3, 4}. The total number of<br />
unordered pairs of disjoint subsets of S is equal to<br />
(A) 25 (B) 34<br />
(C) 42 (D) 41<br />
Ans. [D]<br />
Sol. S = {1, 2, 3, 4}<br />
Possible subsets No. of elements in Ways<br />
Set A Set B<br />
0 0 = 1<br />
1 0 = 4 C 1 = 4<br />
2 0 = 4 C 2 = 6<br />
1 1 = 4 C 2 = 6<br />
3 0 = 4 C 3 = 4<br />
2 1 = 4 C 2 . 2 C 1 = 12<br />
4 0 = 4 C 4 = 1<br />
3 1<br />
4!<br />
= = 4<br />
3! 1!<br />
2 2<br />
4!<br />
= = 3<br />
2! 2! 2!<br />
Total ⇒ 1 + 4 + 6 + 6 + 4 + 12 + 1 + 4 + 3 = 41<br />
24. Let f be a real-valued function defined on the<br />
interval (–1, 1) such that e –x f(x) = 2 +<br />
Ans.<br />
x<br />
4<br />
∫<br />
t + 1 dt , for all x ∈ (–1, 1) , and let f –1 be the<br />
0<br />
inverse function of f. Then (f –1 )′ (2) is equal to<br />
(A) 1 (B) 3<br />
1<br />
(C) 2<br />
1<br />
[B]<br />
Sol. f(x) = 2e x + e x ∫<br />
t + 1 dt<br />
x = 0 ; f(0) = 2<br />
x<br />
0<br />
x<br />
4<br />
(D) e<br />
1<br />
f ′(x) = 2e x + e x 4<br />
∫<br />
t + 1 dt + e x 1+ x<br />
4<br />
0<br />
f ′(0) = 2 + 1 = 3 (f –1 1 1<br />
)′ (2) = =<br />
f ′(0)<br />
3<br />
25. For r = 0, 1, …, 10, let A r , B r and C r denote,<br />
respectively, the coefficient of x r in the<br />
expansions of (1 + x) 10 , (1 + x) 20 and (1+ x) 30 .<br />
10<br />
Then ∑ Ar (B10Br<br />
− C10Ar<br />
) is equal to<br />
r=<br />
1<br />
2<br />
(A) B 10 – C 10 (B) A 10 ( B − C )<br />
10 10A10<br />
⎛ π ⎞<br />
cos ⎜ − α⎟ =<br />
⎝ 2 ⎠<br />
AB.AD<br />
| AB || AD |<br />
8 17<br />
sin α = ⇒ cos α = 9 9<br />
=<br />
40 8 =<br />
3(15) 9<br />
Ans.<br />
Sol.<br />
(C) 0 (D) C 10 – B 10<br />
[D]<br />
A r = 10 C r , B r = 20 C r , C r = 30 C r<br />
10<br />
∑<br />
r=<br />
1<br />
10 C ( 20 C 10 20 C r – 30 C 10 10 C r )<br />
r<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 92<br />
MAY 2010
10<br />
= 20 C 10 ∑<br />
10 Cr<br />
20 C 20–r – 30 10 10<br />
C 10 ∑ Cr.<br />
Cr<br />
r=<br />
1<br />
10<br />
r=<br />
1<br />
= 20 C 10 [ 30 C 20 – 10 C 0 20 C 20 ] – 30 C 10 [ 20 C 10 – ( 10 C 0 ) 2 ]<br />
= 20 C 10 30 C 20 – 20 C 10 – 30 C 10 20 C 10 + 30 C 10<br />
= 30 C 10 – 20 C 10<br />
= C 10 – B 10<br />
SECTION – II<br />
Integer type<br />
This section contains 5 questions. The answer to each<br />
question is a single-digit integer, ranging from 0 to 9.<br />
The correct digit below the question no. in the ORS is<br />
to be bubbled.<br />
26. Let a 1 , a 2 , a 3 , …….., a 11 be real numbers<br />
satisfying a 1 = 15, 27 –2a 2 > 0 and a k =2a k–1 – a k–2<br />
for k = 3, 4, …., 11.<br />
2 2<br />
2<br />
a1 + a 2 + ..... + a11<br />
If<br />
= 90, then the value of<br />
11<br />
a1 + a 2 + ..... + a11<br />
is equal to<br />
11<br />
Ans. [0]<br />
a k + a k−2<br />
Sol. Q a k–1 =<br />
2<br />
2 2<br />
2<br />
a1 + a 2 + ..... + a11<br />
so<br />
= 90<br />
11<br />
⇒ Σ(a + (r –1) d) 2 = 11 × 90<br />
⇒ Σ(a 2 + 2ad (r –1) + (r –1) 2 d 2 ) = 11 × 90<br />
11a 2 10×11 10 × 11×<br />
21<br />
+ 2ad +<br />
d 2 = 11 × 90<br />
2 6<br />
so on solving d = –3<br />
a1 + a 2 + ...... + a11<br />
so<br />
11<br />
11 1<br />
= . . (2 × a1 + (11 –1) (–3))<br />
2 11<br />
= 2<br />
1 (30 – 30) = 0<br />
27. Let f be a function defined on R (the set of all real<br />
numbers) such that<br />
f ′(x) = 2010 (x –2009) (x –2010) 2 (x –<strong>2011</strong>) 3<br />
(x –2012) 4 , for all x ∈ R.<br />
If g is a function defined on R with values in the<br />
interval (0, ∞) such that<br />
f(x) = ln {g(x)}, for all x ∈ R, then the number of<br />
points in R at which g has a local maximum is<br />
Ans. [1]<br />
Sol. g(x) = e f(x)<br />
g′(x) = e f(x) f ′ (x)<br />
g′(x) = 0 ⇒ f ′ (x) = 0 ⇒ x = 2009, 2010, <strong>2011</strong>,<br />
2012<br />
<strong>Point</strong>s of local maxima = 2009, ⇒ only one point<br />
28. Let k be a positive real number and let<br />
⎡ 2k −1<br />
⎢<br />
A = ⎢ 2 k<br />
⎢−<br />
2 k<br />
⎣<br />
2<br />
1<br />
2k<br />
k<br />
2 k ⎤<br />
⎥<br />
− 2k⎥<br />
and<br />
−1<br />
⎥<br />
⎦<br />
⎡ 0 2k −1<br />
k ⎤<br />
⎢<br />
⎥<br />
B = ⎢1<br />
− 2k 0 2 k ⎥<br />
⎢ − k − 2 k 0 ⎥<br />
⎣<br />
⎦<br />
If det (adj A) + det(adj B) = 10 6 , then [k] is equal<br />
to<br />
[Note : adj M denotes the adjoint of a square<br />
matrix M and [k] denotes the largest integer less<br />
than or equal to k]<br />
Ans. [4]<br />
Sol. det (A) =<br />
2k −1<br />
2<br />
− 2<br />
k<br />
k<br />
2<br />
1<br />
2k<br />
k<br />
2 k<br />
− 2k<br />
−1<br />
= (2k –1) [–1 + 4k 2 ] –2 k[–2 k –4k k]<br />
+ 2 k [4k k + 2 k ]<br />
det (A) = (2k –1) (4k 2 –1) + 4 k (2k + 1) + 4k (2k +1)<br />
= (2k –1) (4k 2 –1) + 8k (2k + 1)<br />
det (B) = 0<br />
det (adj A) = (det A) 2 = 10 6 det A = 10 3<br />
8k 3 + 1 –2k –4k 2 + 16k 2 + 8k = 10 3<br />
8k 3 + 12k 2 + 6k – 999 = 0<br />
k = 2 → 64 + 48 + 12 – 999 < 0<br />
k = 3 → 8(27) + 109 + 18 – 999 < 0<br />
k = 4 → 8(64) + 12 (16) + 24 – 999<br />
512 + 192 + 24 – 999 < 0<br />
k = 5 → 8(125) + 12 (25) + 6(5) – 999 > 0<br />
so [k] = 4<br />
29. Two parallel chords of a circle of radius 2 are at a<br />
distance 3 + 1 apart. If the chords subtend at the<br />
π 2π<br />
center, angles of and , where k > 0, then<br />
k k<br />
the value of [k] is<br />
[Note : [k] denotes the largest integer less than or<br />
equal to k]<br />
Ans. [3]<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 93<br />
MAY 2010
Sol.<br />
π/k<br />
π/2k<br />
d = 2 cos k<br />
π + 2 cos<br />
π<br />
2k<br />
3 + 1 3π 3π 3 + 1 1<br />
= cos cos ⇒ cos cos = .<br />
4 4k<br />
4π<br />
k 4π<br />
k 4k<br />
2 2 2<br />
π π =<br />
4k 12<br />
4k = 12<br />
k = 3<br />
30. Consider a triangle ABC and let a, b and c denote<br />
the lengths of the sides opposite to vertices A, B<br />
and C respectively. Suppose a = 6, b = 10 and the<br />
area of the triangle is 15 3 . If ∠ACB is obtuse<br />
and if r denotes the radius of the in-circle of the<br />
triangle, then r 2 is equal to<br />
Ans. [3]<br />
Sol.<br />
B 6 C<br />
∆ = 2<br />
1 ab sin C<br />
10<br />
15 3 = 2<br />
1 6(10) sin C ⇒ sin C = 3 /2<br />
⇒ C = 120°<br />
2<br />
100 + 36 − c<br />
cos C =<br />
⇒ C 2 = 136 + 120° (1/2)<br />
2.10.6<br />
⇒ C 2 = 196 ⇒ C =14<br />
s = 15<br />
r = s<br />
∆ = 3<br />
r 2 = 3<br />
SECTION – III<br />
Paragraph Type<br />
This section contains 2 paragraphs. Based upon each of<br />
the paragraphs 3 multiple choice questions have to be<br />
answered. Each of these question has four choices (A),<br />
(B), (C) and (D) out of which ONLY ONE is correct.<br />
A<br />
Paragraph for Questions No. 31 to 33<br />
Consider the polynomial f(x) = 1 + 2x + 3x 2 +<br />
4x 3 . Let s be the sum of all distinct real roots of<br />
f(x) and let t = |s|.<br />
31. The real number s lies in the interval<br />
⎛ 1 ⎞<br />
⎛ 3 ⎞<br />
(A) ⎜ − , 0 ⎟ (B) ⎜−11<br />
, − ⎟<br />
⎝ 4 ⎠ ⎝ 4 ⎠<br />
⎛ 3 1 ⎞<br />
⎛ 1 ⎞<br />
(C) ⎜−<br />
, − ⎟ (D) ⎜0<br />
, ⎟<br />
⎝ 4 2 ⎠ ⎝ 4 ⎠<br />
Ans. [C]<br />
Sol. f(x) = 4x 3 + 3x 2 + 2x + 1<br />
f ′(x) = 12x 2 + 6x + 2 is always positive<br />
1<br />
f(0) = 1, f(–1/2) = 1/4, f(–3/4) = – 2<br />
Q<br />
⎛ − 3 −1⎞<br />
so root ∈ ⎜ , ⎟<br />
⎝ 4 2 ⎠<br />
the equation have only one real root so<br />
⎛ − 3 −1⎞<br />
⎛ 1 3 ⎞<br />
s ∈ ⎜ , ⎟ and t ∈ ⎜ , ⎟<br />
⎝ 4 2 ⎠ ⎝ 2 4 ⎠<br />
32. The area bounded by the curve y = f(x) and the<br />
lines x = 0, y = 0 and x = t, lies in the interval<br />
⎛ 3 ⎞<br />
⎛ 21 11 ⎞<br />
(A) ⎜ , 3⎟ (B) ⎜ , ⎟<br />
⎝ 4 ⎠ ⎝ 64 16 ⎠<br />
Ans.<br />
Sol.<br />
⎛ 21⎞<br />
(C) (9, 10) (D) ⎜0<br />
, ⎟<br />
⎝ 64 ⎠<br />
[A]<br />
t<br />
⎛<br />
4<br />
⎞<br />
A(t) =<br />
∫<br />
f (x)d(x) = t 4 + t 3 + t 2 + t = t ⎜<br />
1−<br />
t<br />
⎟<br />
0<br />
⎝ 1−<br />
t ⎠<br />
⎛ 175 ⎞<br />
A(1/2) = 15/16 & A (3/4) = 3 ⎜ ⎟<br />
⎝ 256 ⎠<br />
⎛ 3 ⎞<br />
So A(t) ∈ ⎜ , 3⎟ ⎝ 4 ⎠<br />
33. The function f ′(x) is<br />
(A) increasing in<br />
in<br />
⎛<br />
⎜ −<br />
⎝<br />
1 ,<br />
4<br />
⎞ t ⎟<br />
⎠<br />
(B) decreasing in<br />
in<br />
⎛<br />
⎜ −<br />
⎝<br />
1 ,<br />
4<br />
⎞ t ⎟<br />
⎠<br />
⎛ 1 ⎞<br />
⎜−<br />
t,<br />
− ⎟ and decreasing<br />
⎝ 4 ⎠<br />
⎛ 1 ⎞<br />
⎜−<br />
t,<br />
− ⎟ and increasing<br />
⎝ 4 ⎠<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 94<br />
MAY 2010
(C) increasing in (–t, t)<br />
(D) decreasing in (–t, t)<br />
Ans. [B]<br />
Sol. f ′′(x) = 6 (4x + 1)<br />
Paragraph for questions No. 34 to 36<br />
Tangents are drawn from the point P(3, 4) to the<br />
2 2<br />
x y<br />
ellipse + = 1, touching the ellipse at points<br />
9 4<br />
A and B.<br />
34. The coordinates of A and B are<br />
(A) (3, 0) and (0, 2)<br />
⎛ ⎞<br />
(B) ⎜<br />
8 2 161<br />
⎟<br />
⎞<br />
− , and<br />
⎜<br />
⎛ 9 8 − , ⎟<br />
⎝ 5 15 ⎠ ⎝ 5 5 ⎠<br />
Ans.<br />
Sol.<br />
⎛ ⎞<br />
(C) ⎜<br />
8 2 161 − , ⎟ and (0, 2)<br />
⎝ 5 15 ⎠<br />
⎞<br />
(D) (3, 0) and ⎜<br />
⎛ 9 8 − , ⎟<br />
⎝ 5 5 ⎠<br />
[D]<br />
Equation of tangent<br />
y = mx ± 9m<br />
2 + 4<br />
as it passes through (3, 4)<br />
so 4 = 3m ± 9m<br />
2 + 4<br />
m = 2<br />
1 and undefined.<br />
So equation of the tangents will be<br />
x –2y + 5 = 0 and x = 3<br />
⎛ − 9 8 ⎞<br />
so point of contacts are (3, 0) and ⎜ , ⎟<br />
⎝ 5 5 ⎠<br />
35. The orthocentre of the triangle PAB is<br />
⎛ 8 ⎞<br />
⎛ 7 25 ⎞<br />
(A) ⎜5 , ⎟ (B) ⎜ , ⎟<br />
⎝ 7 ⎠ ⎝ 5 8 ⎠<br />
Ans.<br />
Sol.<br />
⎛11<br />
8 ⎞<br />
(C) ⎜ , ⎟<br />
⎝ 5 5 ⎠<br />
[C]<br />
A(3,0)<br />
H<br />
P(3,4)<br />
θ<br />
⎛<br />
(D) ⎜<br />
⎝<br />
⎛ − 9 8 ⎞<br />
B⎜<br />
, ⎟<br />
⎝ 5 5 ⎠<br />
8<br />
25<br />
7 ⎞<br />
, ⎟<br />
5 ⎠<br />
Equation of two altitudes PH and AQ are<br />
3x – y – 5 = 0 and 2x + y – 6 = 0 respectively<br />
⎛11<br />
8 ⎞<br />
so orthocentre will be ⎜ , ⎟<br />
⎝ 5 5 ⎠<br />
36. The equation of the locus of the point whose<br />
distances from the point P and the line AB are<br />
equal, is<br />
(A) 9x 2 + y 2 – 6xy –54 x – 62 y + 241 = 0<br />
(B) x 2 + 9y 2 + 6xy –54x + 62 y –241 = 0<br />
(C) 9x 2 + 9y 2 –6xy –54 x –62 y – 241 = 0<br />
(D) x 2 + y 2 –2xy + 27x + 31y – 120 = 0<br />
Ans. [A]<br />
Sol. Equation of AB is x + 3y – 3 = 0<br />
so required locus will be<br />
2<br />
(x – 3) 2 + (y –4) 2 (x + 3y – 3)<br />
=<br />
10<br />
⇒ 9x 2 + y 2 – 6xy –54x –62y + 241 = 0<br />
SECTION – IV<br />
Matrix Type<br />
This section contains 2 questions. Each question has<br />
four statements (A, B, C and D)given in Column I and<br />
five statements (p, q, r, s and t) in column-II. Any given<br />
statement in Column I can have correct matching with<br />
one or more statement(s) given in Column-II. For<br />
example, if for a given question, statement B matches<br />
with the statements given in q and r, then for that<br />
particular question, against statement B, darken the<br />
bubbles corresponding to q and r in the ORS.<br />
37. Match the statements in Column-I with those in<br />
Column-II.<br />
[Note : Here z takes values in the complex plane<br />
and lm z and Re z denote, respectively, the<br />
imaginary part and the real part of z]<br />
COLUMN-I<br />
COLUMN-II<br />
(A) The set of points (p) an ellipse with<br />
z satisfying<br />
4<br />
eccentricity 5<br />
|z – i|z|| = |z + i|z||<br />
is contained in or e<br />
qual to<br />
(q) the set of points z<br />
satisfying Im z = 0<br />
(B) The set of points z (r) the set of points z<br />
Satisfying satisfying |Im z| ≤ 1<br />
|z + 4| + |z –4| = 10<br />
is contained in or equal to<br />
(C) If |w| = 2, then the set (s) the set of points z<br />
of points satisfying |Re z| ≤ 2<br />
z = w – w<br />
1 is contained<br />
in or equal to<br />
(D) If |w| =1, then the set (t) the set of points z<br />
of points satisfying |z| ≤ 3<br />
z = w + w<br />
1 is contained<br />
in or equal to<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 95<br />
MAY 2010
Ans.<br />
Sol.<br />
A → q, r ; B → p ; C → p, s, t ; D → q, r, s, t<br />
(A) Let |Z| = r ∀ r ∈ R<br />
Z − ir<br />
= 1Which is the equation of line of<br />
Z + ir<br />
perpendicular<br />
bisector of y = r & y = – r that is y = 0<br />
(B) |Z + 4| + |Z – 4| = 10<br />
it will represent on ellipse<br />
having foci (–4, 0), (4, 0)<br />
2 2<br />
x y<br />
so its equation will be + = 1<br />
25 9<br />
whose eccentricity is 4/5<br />
(C) Let w = 2e iθ .<br />
z = 2<br />
3 cos + 2<br />
5 i sin θ<br />
(D) Let w = e iθ<br />
Z = e iθ + e –iθ<br />
= 2 cos θ.<br />
38. Match the statements in Column-I with the<br />
values in Column-II.<br />
COLUMN-I<br />
COLUMN-II<br />
(A) A line from the origin (p) –4<br />
meets the lines<br />
x − 2 y z + 1<br />
= and<br />
21<br />
Ans.<br />
1<br />
8<br />
x −<br />
3<br />
2<br />
=<br />
−− =<br />
1<br />
y<br />
3 =<br />
−+<br />
1 1<br />
z −1<br />
at P and Q respectively.<br />
If length PQ = d, then d 2 is<br />
(B) The values of x satisfying (q) 0<br />
tan –1 (x + 3) – tan –1 (x –3)<br />
= sin –1 ⎛ 3 ⎞<br />
⎜ ⎟ are,<br />
⎝ 5 ⎠<br />
(C) Non-zero vectors a r , b r (r) 4<br />
and c r satis1fy a r . b r = 0,<br />
(b r – a r ). ( b r + c r ) = 0 and<br />
2 | b r + c r |= | b r – a r |.<br />
If a r = µ b r + 4 c r , then the<br />
possible values of µ are<br />
(D) Let f be the function on (s) 5<br />
[–π, π] given by<br />
f(0) = 9 and f(x)<br />
⎛ 9x ⎞ ⎛ x ⎞<br />
= sin ⎜ ⎟ sin⎜<br />
⎟ for x ≠ 0<br />
⎝ 2 ⎠ ⎝ 2 ⎠<br />
π<br />
2<br />
The value of π ∫<br />
f (x)dx is<br />
−π<br />
(t) 6<br />
A → t ; B → p, r ; C → q, s ; D → r<br />
Sol.<br />
(A) Let P ≡ (λ + 2, 1 –2 λ, λ – 1)<br />
Q ≡ (2µ + 3<br />
2 ; – µ –3, µ + 1)<br />
equation line PQ<br />
→<br />
r = (λ + 2) î + (1 – 2λ) ĵ + (λ – 1) kˆ<br />
+ α ((2µ – λ + 3<br />
2 ) î + (2λ – µ – 4) ĵ<br />
+ (µ + 2 – λ) kˆ<br />
∴ This line passing through origin so.<br />
λ + 2 + α (2µ − λ + 3<br />
2 ) = 0<br />
1 – 2λ + α(2λ – µ – 4) = 0<br />
λ – 1 + α(µ – λ + 2) = 0<br />
on solving above three µ = 3<br />
1 & λ = 3<br />
10 −10 4<br />
So P ≡ (5, – 5, 2) & Q ≡ ( , , )<br />
3 3 3<br />
So PQ = 6 ⇒ (PQ) 2 = 6<br />
(B) tan –1 (x + 3) – tan –1 (x –3) = sin –1 3<br />
5<br />
tan –1 6<br />
= tan –1 3<br />
x<br />
2 − 8 4<br />
⇒ x 2 – 8 = 8<br />
⇒ x 2 = 16 ⇒ x = ± 4<br />
(C) | b r | 2 + b r . c r = a r . c r …. (1)<br />
put a r = µ b r + 4 c r ∀ a r . b r = 0 ⇒ b r . c r = – 4<br />
µ | b<br />
r<br />
|<br />
2<br />
from (1) and (2)<br />
2<br />
b 16<br />
2 =<br />
2<br />
…(2)<br />
…(3)<br />
c 4 − µ + µ<br />
∴ 2| b r + c r | = | b r –a r | and a r = µ b r + 4 c r<br />
2<br />
b 12<br />
2 =<br />
2<br />
c<br />
3 − 2µ + µ<br />
from (3) and (4)<br />
m = 0,5<br />
9x<br />
sin<br />
(D) f(x) =<br />
2<br />
=<br />
x<br />
sin<br />
2<br />
π<br />
2<br />
I = π ∫<br />
f (x)dx<br />
−π<br />
sin 5x<br />
+<br />
sin x<br />
sin 4x<br />
sin x<br />
… (4)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 96<br />
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π<br />
4<br />
= π ∫<br />
f (x)dx<br />
0<br />
π<br />
4 sin 5x<br />
= π ∫ sin x<br />
0<br />
π/<br />
2<br />
= π<br />
8<br />
∫<br />
0<br />
π/<br />
2<br />
= π<br />
8<br />
∫<br />
= 4<br />
0<br />
sin 5x<br />
dx<br />
sin x<br />
sin (3x + 2x)<br />
dx<br />
sin x<br />
π/<br />
2<br />
8<br />
= π ∫<br />
( 1+<br />
2cos 4x)dx<br />
0<br />
Total force =<br />
π/<br />
2<br />
∫<br />
2<br />
0 0<br />
σ<br />
πR<br />
2ε<br />
σ<br />
sin 2θdθ<br />
=<br />
2ε<br />
2<br />
2<br />
0<br />
× πR<br />
40. A block of mass 2 kg is free to move along the x-<br />
axis. It is at rest and from t = 0 onwards it is<br />
subjected to a time dependent force F(t) in the x<br />
direction. The force F(t) varies with t as shown in<br />
the figure. The kinetic energy of the block after<br />
4.5 seconds is -<br />
F(t)<br />
4N<br />
2<br />
PHYSICS<br />
O<br />
3s<br />
4.5s<br />
t<br />
SECTION – I<br />
Single Correct Choice Type<br />
This section contains 6 multiple choice questions. Each<br />
question has 4 choices (A), (B), (C) and (D), out of<br />
which ONLY ONE is correct.<br />
39. A uniformly charged thin spherical shell of radius<br />
R carries uniform surface charge density of σ per<br />
unit area. It is made of two hemispherical shells,<br />
held together by pressing them with force F (see<br />
figure). F is proportional to -<br />
Ans.<br />
Sol.<br />
F<br />
(A)<br />
1 σ<br />
ε<br />
0<br />
2<br />
2<br />
R<br />
1 σ<br />
(C)<br />
ε0<br />
R<br />
[A]<br />
θ<br />
dθ<br />
2<br />
2<br />
σ<br />
2ε<br />
0<br />
(B)<br />
(D)<br />
dA = 2πR sin θ × Rdθ<br />
2<br />
ε<br />
1 2<br />
0<br />
σ<br />
1 σ<br />
ε<br />
2<br />
R<br />
2<br />
0 R<br />
F<br />
(Electrostatic pressure)<br />
σ<br />
dF = × dA<br />
2ε0<br />
Component of dF along vertical axis = dF cos θ<br />
Ans.<br />
Sol.<br />
(A) 4.50 J<br />
(C) 5.06 J<br />
[C]<br />
F<br />
4N<br />
–2N<br />
O 3<br />
(B) 7.50 J<br />
(D) 14.06 J<br />
4.5<br />
t (sec)<br />
4<br />
m = −<br />
3<br />
→<br />
At t = 4.5 sec F = – 2N<br />
⎡1<br />
⎤ ⎡1<br />
⎤<br />
Total Impulse I = ⎢ × 3×<br />
4⎥ –<br />
⎣ 2<br />
⎢ × 2×<br />
1. 5⎥ ⎦ ⎣ 2 ⎦<br />
⇒ = 6 – 1.5 = 4.5 SI unit<br />
Impulse = change in momentum<br />
4.5 = 2[v – 0]<br />
4.5<br />
v = = 2.25 m/sec<br />
2<br />
1<br />
K.E. = × 2 × (2.25) 2 = 5.06 J<br />
2<br />
41. A tiny spherical oil drop carrying a net charge q is<br />
balanced in still air with a vertical uniform<br />
81 π 5<br />
electric field of strength × 10 Vm –1 . When<br />
7<br />
the field is switched off, the drop is observed to<br />
fall with terminal velocity<br />
2 × 10 –3 ms –1 . Given g = 9.8 m s –2 , viscosity of<br />
the air = 1.8 × 10 –5 Nsm –2 and the density of oil =<br />
900 kg m –3 , the magnitude of q is -<br />
(A) 1.6 × 10 –19 C (B) 3.2 × 10 –19 C<br />
(C) 4.8 × 10 –19 C (D) 8.0 × 10 –19 C<br />
Ans.<br />
[D]<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 97<br />
MAY 2010
Sol.<br />
qE = mg<br />
⎡81π 5 ⎤ 4<br />
⇒ q ⎢ × 10 ⎥ = 900 × πr 3 × 9.8<br />
⎣ 7 ⎦ 3<br />
3<br />
900×<br />
4×<br />
r × 9.8×<br />
7<br />
q =<br />
...... (1)<br />
5<br />
3×<br />
81×<br />
10<br />
v T = 2 × 10 –3 m/sec<br />
2<br />
2 × 10 –3 2 r × 900×<br />
9.8<br />
= × 9<br />
−5<br />
1.8×<br />
10<br />
−5<br />
r 2 18×<br />
1.8×<br />
10 × 10<br />
=<br />
2×<br />
900×<br />
9.8<br />
10 = 18.36 × 10 –12<br />
−3<br />
= 0.1836 × 10 –<br />
r = 4.284 × 10 –6 m<br />
3600×<br />
9.8×<br />
7<br />
q =<br />
5<br />
243×<br />
10<br />
× 78.62 × 10 –18<br />
q = 0.799 × 10 –18 ≈ 8 × 10 –19 C<br />
42. A hollow pipe of length 0.8 m is closed at one<br />
end. At its open end a 0.5 m long uniform string<br />
is vibrating in its second harmonic and it<br />
resonates with the fundamental frequency of the<br />
pipe. If the tension in the wire is 50 N and the<br />
speed of sound is 320 ms –1 , the mass of the string<br />
is -<br />
(A) 5 grams<br />
(B) 10 grams<br />
(C) 20 grams<br />
(D) 40 grams<br />
Ans. [B]<br />
v<br />
Sol. Fundamental frequency of closed pipe =<br />
4l<br />
320<br />
⇒ =<br />
4×<br />
0.8<br />
320 = 100 Hz<br />
3.2<br />
Frequency of 2 nd v 1 T<br />
Harmonic of string = = l l µ<br />
100 =<br />
⇒ 100 =<br />
1 50<br />
l m / l<br />
50<br />
m×<br />
0.5<br />
⇒ 100 =<br />
100<br />
m<br />
100<br />
10000 = ⇒ m = 10<br />
–2<br />
kg = 10 gm<br />
m<br />
43. A biconvex lens of focal length 15 cm is in front<br />
of a plane mirror. The distance between the lens<br />
and the mirror is 10 cm. A small object is kept at<br />
a distance of 30 cm from the lens. The final image<br />
is -<br />
(A) virtual and at a distance of 16 cm from the<br />
mirror<br />
(B) real and at a distance of 16 cm from the<br />
mirror<br />
(C) virtual and at a distance of 20 cm from the<br />
mirror<br />
(D) real and at a distance of 20 cm from the<br />
mirror<br />
Ans.<br />
Sol.<br />
[B]<br />
30cm<br />
6 cm<br />
10 cm<br />
10cm<br />
20 cm<br />
Refraction of reflected light by lens<br />
f = + 15 cm<br />
u = + 10 cm<br />
1 1 1 1 1 1<br />
– = ⇒ − =<br />
v u f v 10 15<br />
v = 6 cm<br />
as incident rays are converging so refracted rays<br />
will converge more and final image is real.<br />
44. A vernier calipers has 1 mm marks on the main<br />
scale. It has 20 equal divisions on the vernier<br />
scale which match with 16 main scale divisions.<br />
For this vernier calipers, the least count is -<br />
(A) 0.02 mm<br />
(B) 0.05 mm<br />
(C) 0.1 mm<br />
(D) 0.2 mm<br />
Ans.<br />
Sol.<br />
[D]<br />
Least Count = M.S. Reading – V.S. Reading<br />
.... (1)<br />
20 V.S. = 16 M.S. or 16 mm<br />
16 16<br />
1 V.S. = M.S. or mm<br />
20 20<br />
In equation (1)<br />
⎛ 16 ⎞<br />
Least Count = ⎜1 − ⎟ mm<br />
⎝ 20 ⎠<br />
= 0.2 mm<br />
SECTION – II<br />
Integer Type<br />
This section contains Five questions. The answer to<br />
each of the questions is a single-digit integer, ranging<br />
from 0 to 9. The correct digit below the question<br />
number in the ORS is to be bubbled.<br />
45. A large glass slab (µ = 5/3) of thickness 8 cm is<br />
placed over a point source of light on a plane<br />
surface. It is seen that light emerges out of the top<br />
surface of the slab from a circular area of radius R<br />
cm. What is the value of R ?<br />
Sol. [6]<br />
3<br />
sin θ cr = 5<br />
⇒ tan θ cr = 4<br />
3<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 98<br />
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8 cm<br />
R = 8 tan θ cr<br />
R<br />
47. To determine the half life of a radioactive<br />
dN(t)<br />
element, a student plots a graph of l n<br />
θ<br />
dt<br />
cr<br />
versus t. Here (t)<br />
is the rate of radioactive<br />
dt<br />
decay at time t. If the number of radioactive<br />
nuclei of this element decreases by a factor of p<br />
after 4.16 years, the value of p is -<br />
6<br />
5<br />
50 m in 30<br />
dN(t) 4<br />
ln<br />
dt 3<br />
2<br />
1<br />
2 3 4 5 6 7 8<br />
Years<br />
Ans. 8<br />
Sol. From graph<br />
1<br />
slope = = 0.5 year<br />
–1<br />
2<br />
dN = Ne<br />
–λt<br />
dt<br />
⎛ dN ⎞<br />
ln ⎜ ⎟<br />
⎝ dt ⎠<br />
= ln (N) – λt<br />
so comparing we get λ = 0.5 year –1<br />
= 0.693<br />
t 1/2<br />
0.5<br />
year<br />
t = 4.16 years<br />
4.16<br />
so No. of half lives =<br />
0.693<br />
u 1<br />
× 0.5 = 3<br />
N<br />
N 0 → 0 N<br />
→ 0 N<br />
→ 0<br />
2 4 8<br />
⇒ p = 8<br />
48. A diatomic ideal gas is compressed adiabatically<br />
to<br />
1 of its initial volume. In the initial<br />
32<br />
temperature of the gas is T i (in Kelvin) and the<br />
final temperature is aT i , the value of a is -<br />
Ans. [4]<br />
Sol. for adiabatic process<br />
TV γ–1 = const.<br />
7<br />
7<br />
1<br />
⇒ T i V<br />
5<br />
1<br />
⎛ V ⎞ 5<br />
= aT i ⎜ ⎟<br />
u 2<br />
⎝ 32 ⎠<br />
⇒ a = 4<br />
50 − 25 25<br />
= m/sec<br />
30 30<br />
= 8 × 4<br />
3 = 6 cm<br />
46. Image of an object approaching a convex mirror<br />
of radius of curvature 20 m along its optical axis<br />
Ans. [3]<br />
Sol.<br />
25<br />
is observed to move from m to 3 7<br />
seconds. What is the speed of the object in km<br />
per hour ?<br />
For position of object initially when image<br />
25<br />
was at m<br />
3<br />
1 3 1<br />
– = − + 10 25 u<br />
3 1 1 – =<br />
25 10 u<br />
12 –10 1<br />
= u<br />
100<br />
u 1 = 50<br />
O<br />
50<br />
For position of object when image is at m 7<br />
1 7 1<br />
– = − + 10 50 u<br />
7 1 1 – =<br />
50 10 u<br />
u 2 = 25<br />
O<br />
Speed of object =<br />
25<br />
= × 30 1000<br />
49. At time t = 0, a battery of 10 V is connected<br />
across points A and B in the given circuit. If the<br />
capacitors have no charge initially, at what time<br />
(in seconds) does the voltage across them become<br />
4 V ?<br />
(Take : ln 5 = 1.6, ln 3 = 1.1]<br />
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2MΩ<br />
2µF<br />
Sol.<br />
F T<br />
A<br />
Ans. [2]<br />
2MΩ<br />
2×10 6<br />
2×10 –6<br />
2µF<br />
B<br />
r<br />
R<br />
Sol.<br />
A<br />
2×10 6 B<br />
2×10 –6<br />
10 6 4×10 –6<br />
10V<br />
F T = 2πrT<br />
Net vertically upward force<br />
⎛ r ⎞ 2πr<br />
2 T<br />
⇒ 2πrT ⎜ ⎟ =<br />
⎝ R ⎠ R<br />
q = CV 0 (1 – e –t/RC )<br />
V = V 0 (1 – e –t/RC )<br />
4 = 10(1 – e –t/4 )<br />
3 = 5e –t/4<br />
t<br />
log3 = log 5 – 4<br />
1.1 – 1.6 = – 4<br />
t<br />
⇒ t = 2 sec<br />
SECTION – III<br />
Paragraph Type<br />
This section contains 2 paragraphs.. Based upon each<br />
of paragraph 3 multiple choice question have to be<br />
answered. Each of these questions has four choices (A),<br />
(B), (C) and (D) for its answer, out of which ONLY<br />
ONE is correct.<br />
Paragraph for Questions No. 50 to 52<br />
When liquid medicine of density ρ is to be put in<br />
the eye, it is done with the help of a dropper. As<br />
the bulb on the dropper is pressed, a drop forms at<br />
the opening of the dropper. We wish to estimate<br />
the size of the drop. We first assume that the drop<br />
formed at the opening is spherical because that<br />
requires a minimum increase in its surface energy.<br />
To determine the size, we calculate the net<br />
vertical force due to the surface tension T when<br />
the radius of the drop is R. When this force<br />
becomes smaller than the weight of the drop, the<br />
drop gets detached from the dropper.<br />
50. If the radius of the opening of the dropper is r, the<br />
vertical force due to the surface tension on the<br />
drop of radius R (assuming r
Ans.<br />
Sol.<br />
[D]<br />
Bohr quantization principle<br />
nh nh<br />
L = = Iω ⇒ ω =<br />
2π<br />
2πI<br />
2<br />
1 2 1 ⎛ nh ⎞<br />
Rotational KE = I ω = I⎜<br />
⎟ =<br />
2 2 ⎝ 2πI<br />
⎠<br />
2<br />
2<br />
n h<br />
2<br />
8π<br />
I<br />
54. It is found that the excitation frequency from<br />
ground to the first excited state of rotation for the<br />
CO moelcule is close to π<br />
4 × 10 11 Hz. Then the<br />
moment of inertia of CO molecule about its<br />
center of mass is close to (Take h = 2π × 10 –34 J s)<br />
(A) 2.76 × 10 –46 kg m 2 (B) 1.87 × 10 –46 kg m 2<br />
(C) 4.67 × 10 –47 kg m 2 (D) 1.17 × 10 –47 kg m 2<br />
Ans. [B]<br />
Sol. ∆E = E 2 – E 1<br />
2 2 2 2 2<br />
2 h 1 h 3h<br />
= – = = hν<br />
2 2 2<br />
8π<br />
I 8π<br />
I 8π<br />
I<br />
56. Two transparent media of refractive indices µ 1<br />
and µ 3 have a solid lens shaped transparent<br />
material of refractive index µ 2 between them as<br />
shown in figures in Column II. A ray traversing<br />
these media is also shown in the figures. In<br />
Column I different relationships between µ 1 , µ 2<br />
and µ 3 are given. Match them to the ray diagrams<br />
shown in Column II.<br />
Column I<br />
(A) µ 1 < µ 2<br />
Column II<br />
(p)<br />
µ 3<br />
µ 2 µ 1<br />
(B) µ 1 > µ 2 (q) µ 3 µ 2 µ 1<br />
(C) µ 2 = µ 3 (r) µ 3 µ 2 µ 1<br />
When ν = π<br />
4 × 10 11 Hz<br />
Solving I = 1.87 ×10 –46 kg – m 2<br />
55. In a CO molecule, the distance between C (mass<br />
= 12 a.m.u.) and O (mass = 16 a.m.u.) where 1<br />
a.m.u. = 3<br />
5 ×10 –27 kg, is close to<br />
(A) 2.4 × 10 –10 m (B) 1.9 × 10 –10 m<br />
(C) 1.3 × 10 –10 m (D) 4.4 × 10 –11 m<br />
Ans. [C]<br />
2 2<br />
Sol. I = m 1r 1 + m 2r 2<br />
Where m 1 = 12 amu<br />
m 2 = 16 amu<br />
m 1 r 1 = m 2 r 2<br />
r 1 + r 2 = r where r → distance between C & O.<br />
Putting and solving<br />
r = 1.279 × 10 –10 m<br />
~_ 1.3 × 10 –10 m<br />
SECTION – IV<br />
Matrix Type<br />
This Section contains 2 questions. Each question has<br />
four statements (A, B, C and D) given in Column I and<br />
five statements (p, q, r, s and t) in Column II. Any<br />
given statement in Column I can have correct<br />
matching with one or more statement(s) given in<br />
Column II. For example, if for a given question,<br />
statement B matches with the statements given in q<br />
and r, then for that particular question, against<br />
statement B, darken the bubbles corresponding to q<br />
and r in the ORS.<br />
(D) µ 2 > µ 3 (s) µ 3 µ 2 µ 1<br />
(t) µ 3 µ 2 µ 1<br />
Ans. [A → p,r; B → q,s,t; C → p,r,t D → q,s]<br />
Sol. For (p) µ 2 > µ 1<br />
as light rays bend towards normal at first refraction<br />
µ 2 = µ 3 as no refraction occurs at second refraction<br />
Option : (A), (C)<br />
For (q)<br />
µ 2 < µ 1 as bend away from normal at first refraction<br />
µ 3 < µ 2 as bends away from normal at second refraction<br />
Option (B), (D)<br />
For (r)<br />
µ 2 > µ 1 as bend towards the normal at first refraction<br />
µ 2 = µ 3 as no refraction occurs at second refraction<br />
Option (A), (C)<br />
For (s)<br />
µ 2 < µ 1 as bend away from normal at first refraction<br />
µ 3 < µ 2 as bend away from normal at second refraction<br />
Option (B), (D)<br />
For (t)<br />
µ 2 < µ 1 as bend away from normal at first refraction<br />
µ 2 = µ 3 as no refraction occurs at second refraction<br />
Option (B), (C)<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 101<br />
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57. You are given many resistances, capacitors and<br />
inductors. These are connected to a variable DC<br />
voltage source (the first two circuits) or an AC<br />
voltage source of 50 Hz frequency (the next three<br />
circuits) in different ways as shown in<br />
Column II. When a current I (steady state for DC<br />
or rms for AC) flows through the circuit, the<br />
corresponding voltage V 1 and V 2 . (indicated in<br />
circuits) are related as shown in Column I. Match<br />
the two<br />
Column I<br />
(A)<br />
I ≠ 0, V 1 is<br />
proportional to I<br />
Column II<br />
(p)<br />
V 1<br />
6mH<br />
V 1<br />
V<br />
V 2<br />
3µF<br />
V 2<br />
V 1 = I X L ; V 2 = IR<br />
So V 2 > V 1<br />
V 2 ∝ I<br />
also V 1 ∝ I Option (A),<br />
(B), (D)<br />
For (s) V 1 = I X L<br />
1<br />
V 2 = I X C where X C =<br />
ωC<br />
1061 Ω<br />
again V 1 ∝ I; V 2 ∝ I, I ≠ 0<br />
Option (A), (B) (D)<br />
For (t) V 1 = IR when R = 1000 Ω<br />
V 2 = I X C when X C 1061 Ω<br />
V 2 > V 1<br />
V 1 , V 2 ∝ I and I ≠ 0<br />
Option (A), (B), (D)<br />
(B) I ≠ 0, V 2 > V 1<br />
(q)<br />
6mH<br />
V 1<br />
2Ω<br />
V<br />
V 2<br />
WHICH IS THE HIGHEST<br />
WATERFALL IN THE WORLD ?<br />
(C) V 1 = 0, V 2 = V<br />
(r)<br />
6mH<br />
2Ω<br />
~ V<br />
V 1<br />
V 2<br />
(D)<br />
I ≠ 0, V 2 is<br />
proportional to I<br />
(s)<br />
6mH<br />
~ V<br />
3µF<br />
V 1<br />
V 2<br />
(t)<br />
1kΩ 3µF<br />
~ V<br />
Ans. [A → r,s,t; B → q,r,s,t; C → p,q; D →<br />
q,r,s,t]<br />
Sol. For (p) Insteady state when I = constant<br />
V L = 0 = V 1<br />
So V 2 = V<br />
Option (C)<br />
For (q) V 1 = 0 again as I = constant<br />
V 2 = V<br />
Also V 2 = IR ⇒ Propotional to I.<br />
Option (B), (C), (D)<br />
For (r) X L = ωL = (100 π) 6 × 10 –3 1.88 Ω<br />
R = 2Ω<br />
The highest waterfall in the world is the<br />
Angel Falls in Venezuela. At a towering height of<br />
979m did you know that each drop of water takes 14<br />
seconds to fall from the top to the bottom. The<br />
water flows from the top of a “Tepui” which is a flat<br />
topped mountain with vertical sides.<br />
The waterfall which despite being known to the local<br />
indians for thousands of years was originally called<br />
the “Churun Meru” but for some reason they were<br />
renamed by an American bush pilot called Jimmy<br />
Angel, who noticed them in 1935 whilst flying over<br />
the area looking for gold.<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 102<br />
MAY 2010
XtraEdge Test Series<br />
ANSWER KEY<br />
<strong>IIT</strong>- <strong>JEE</strong> <strong>2011</strong> (May issue)<br />
PHYSICS<br />
Ques 1 2 3 4 5 6 7 8 9 10<br />
Ans B D C C A B C D A,C A,C,D<br />
Ques 11 12 13 14 15 16 17 18<br />
Ans A,C,D B,C C B D B D D<br />
19 A → R B → P C → T D → S<br />
20 A → S B → P C → Q D → R<br />
CHEMISTRY<br />
Ques 1 2 3 4 5 6 7 8 9 10<br />
Ans D D C D C A A D A,D B,C,D<br />
Ques 11 12 13 14 15 16 17 18<br />
Ans A,B B,C C D A C D D<br />
19 A → P,Q,S B → P,Q,R,S C → P D → Q,R,S,T<br />
20 A → P B → Q C → P,R D → S,P<br />
MATHEMATICS<br />
Ques 1 2 3 4 5 6 7 8 9 10<br />
Ans C C C A B B C A A,B,C A,C,D<br />
Ques 11 12 13 14 15 16 17 18<br />
Ans B C B A B B C C<br />
19 A → P,Q,R,S B → R C → R,S D → Q<br />
20 A → P,R B → P,S C → Q,S D → Q,S<br />
<strong>IIT</strong>- <strong>JEE</strong> 2012 (May issue)<br />
PHYSICS<br />
Ques 1 2 3 4 5 6 7 8 9 10<br />
Ans A A C B D C C B B B,C,D<br />
Ques 11 12 13 14 15 16 17 18<br />
Ans A,B,C,D C,D C A B B C A<br />
19 A → R,T B → S C → P D → Q<br />
20 A → R B → P C → S D → T<br />
CHEMISTRY<br />
Ques 1 2 3 4 5 6 7 8 9 10<br />
Ans A D C D B B C A A,B,C A,B<br />
Ques 11 12 13 14 15 16 17 18<br />
Ans A B A D C D D A<br />
19 A → P B → T C → S D → R<br />
20 A → P,R,S,T B → P C → P,Q,R,T D → S<br />
MATHEMATICS<br />
Ques 1 2 3 4 5 6 7 8 9 10<br />
Ans C B B B A A C B A,C,D A,C<br />
Ques 11 12 13 14 15 16 17 18<br />
Ans A,C,D A,D A C B A B D<br />
19 A → S B → R C → Q D → P<br />
20 A → P B → Q C → P D → R<br />
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 103<br />
MAY 2010
XtraEdge for <strong>IIT</strong>-<strong>JEE</strong> 104<br />
MAY 2010