1. Xtra Edge February 2012 - Career Point

Volume - 7 Issue - 8 

February, 2012 (Monthly Magazine) 

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Editor : Pramod Maheshwari 

Dear Students, 

All of us live and work within fixed patterns. These patterns and habits 

determine the quality of our life and the choices we make in life. There 

are a few vital things to know about ourselves. We should become 

aware of how much we influence others, how productive we are and 

what can help us to achieve our goals. It is important to create an 

environment which will promote our success. We should consciously 

create a system that would enable us to achieve our goals. Most of us 

live in systems which have come our way by an accident, circumstances 

or people we have met over a period of time. We are surrounded by our 

colleagues or subordinates who happened to be there by the fact of 

sheer recruitment earlier or later by the management. Our daily routines 

and schedules have been formed on the basis of convenience, 

coincidence, and the expectations of society and sometimes due to 

superstitions. The trick for success is to have an environment that helps 

in attaining our goals. Control your life. Make an effort to launch your 

day with a great start. A law of physics says that an object set in motion 

tends to remain in motion. It is the same thing with daily routine. To 

have a good start each morning will keep you upbeat during the day. If 

you begin the day stressed, you will tend to remain so that way. The 

best is to create a course of action or conditions where you are not 

hassled for being late for a meeting, worried about household affairs or 

distracted by happenings in the world. 

Aim to be highly successful. Control the direction of your life. Not only 

should you start the day on a cheerful note but also continue to do so 

during the day. Keep yourself stimulated and invigorated during the 

entire day. Start your day with a purpose. Have a daily direction and 

trajectory of action. It will keep you on your course all day long. 

Throughout the day reinforce your positive values and your choices. 

Anything that helps you in maintaining your highest values and your 

most important priorities should be welcome. Be in control of your life 

and work. Create and sustain a wonderful environment filled with 

beauty, peace, inspiration and hope. 

Plan your day in such a way that suits your plans objectives and makes 

you feel just right with the right amount of encouragement during the 

entire day. You should give a direction to your day and timing. 

Presenting forever positive ideas to your success. 

Yours truly 

Pramod Maheshwari, 

B.Tech., IIT Delhi 

Worry is a misuse of imagination. 

Editorial 

XtraEdge for IIT-JEE 1 FEBRUARY 2012

S 

Volume-7 Issue-8 

February, 2012 (Monthly Magazine) 

NEXT MONTHS ATTRACTIONS 

Much more IIT-JEE News. 

Know IIT-JEE With 15 Best Questions of IIT-JEE 

Challenging Problems in Physics,, Chemistry & Maths 

Key Concepts & Problem Solving strategy for IIT-JEE. 

IIT-JEE Mock Test Paper with Solution 

AIEEE & BIT-SAT Mock Test Paper with Solution 

Success Tips for the Months 

• If one asks for success and prepares for 

failure, he will get the situation he has 

prepared for. 

• Loser's visualize the penalties of failure. 

Winner's visualize the rewards of success. 

• Treat others as if they were what they 

ought to be and you help them to become 

what they are capable of being. 

• You never achieve real success unless you 

like what you are doing 

• The first step toward success is taken when 

you refuse to be a captive of the 

environment in which you first find 

yourself. 

• Believe in yourself ! Have faith in your 

abilities ! without a humble but reasonable 

confidence in your own powers you can not 

be successful or happy. 

CONTENTS 

INDEX PAGE 

Regulars .......... 

NEWS ARTICLE 3 

• 25% increase in number of girls from Bombay 

zone for IIT-JEE 

• IIT placements : Companies Back on hiring 

track 

IITian ON THE PATH OF SUCCESS 5 

Dr. Amitabha Ghosh 

KNOW IIT-JEE 6 

Previous IIT-JEE Question 

Study Time........ 

DYNAMIC PHYSICS 14 

8-Challenging Problems [Set # 10] 

Students’ Forum 

Physics Fundamentals 

Matter Waves, Photo-electric Effect 

Thermal Expansion, Thermodynamics 

CATALYSE CHEMISTRY 29 

Key Concept 

Carbonyl Compound 

Co-ordination Compound & 

Metallurgy 

Understanding : Physical Chemistry 

DICEY MATHS 36 

Mathematical Challenges 

Students’ Forum 

Key Concept 

Integration 

Trigonometrical Equation 

Test Time .......... 

XTRAEDGE TEST SERIES 47 

Class XII – IIT-JEE 2012 Paper 

Class XI – IIT-JEE 2013 Paper 

Mock Test-3 (CBSE Board Pattern) [Class # XII] 68 

Solution of Mock Test-2 & 3 (CBSE Pattern) 


25% increase in number of girls 

from Bombay zone for IIT-JEE 

There are far more girls set to appear 

for the Indian Institute of Technology 

Joint Entrance Exam (IIT-JEE) in the 

Bombay zone in April 2012 than 

before. The number of applications by 

girls has increased by nearly 25% 

compared to last year. The Bombay 

zone includes Maharashtra, Goa, 

This year, girls account for nearly 

27% of the total candidates applying, 

with around 20,300 girls from among 

76,600 candidates. The IITs have been 

constantly trying to address the 

skewed male-female ratio at the IITs. 

“This is a significant increase,” said 

AV Mahajan, chairperson for JEE 

2012, Bombay zone. “One reason for 

this could be that it was free for girls 

to apply online this year as well as 

cheaper to apply offline as compared 

to male applicants.” 

Two IT-BHU students to start 

entrepreneurship club for school 

students 

VARANASI: When they cracked IIT- 

JEE two years ago, getting attractive 

jobs in top IT companies was their 

dream. Now, they want to promote 

entrepreneurial skills in young school 

children (Class X onwards) by 

establishing entrepreneurship club for 

schools. 

Bhanu Swami and Varun Agrawal, B 

Tech (III) students of electrical 

engineering of Institute of 

Technology, Banaras Hindu 

University (IT-BHU), are on their way 

to launch country's first entrepreneurship 

club targeting students (Class 

X onwards). The initiative not only 

aims at tapping and promoting 

entrepreneurial skills in young 

students, but also promises to turn 

these students into entrepreneurs. This 

way they would become job providers 

rather than job seekers. 

Akash 2 tablet to be launched by 

April 2012: Govt. 

The government has expressed 

confidence it will be successful in 

bringing out the improved version of 

the Aakash tablet by April this year. 

The Aakash 2 is likely to have several 

improved specifications such as 

elongated battery life and faster 

processor. The government has been 

striving hard to make its Aakash project 

successful. The device, which is also 

touted as the world's cheapest tablet, 

faced harsh criticism from several 

quarters for its poor quality and dismal 

features. 

IIT Rajasthan, which developed the 

prototype of the device along with Data 

Wind, found a series of faults in the 

device, prompting the government to 

reconsider extension of the LC to the 

Canadian company. Under pressure to 

provide better endowed low-cost 

device, the government has now its 

turned attention to the Aakash 2, which 

is likely to come at the same price tag 

of the original Aakash. 

HRD Minister Kapil Sibal has said that 

the government will be looking for 

more manufacturers to manufacture the 

Aakash 2 tablets, attributing 

the massive demand for the move. 

“We want to make sure that the 

upgraded product caters to the need of 

the customers... We have involved ITI 

in order to upgrade it... We will be able 

to bring in Aakash-II by April," Kapil 

Sibal said. 

Last week media reports 

had hinted that the government might 

shelve its Aakash project in view of the 

harsh criticism from most quarters. The 

reports were triggered by DataWind's 

opposition to certain test criteria 

suggested by the IIT. Read our previous 

coverage 

IIT Techfest 2012 : All about 

Robotics 

One of the highlights of the Techfest 

2012, which held at IIT Bombay, 

was its international exhibitions 

arena. Evidently dominated by 

robots, the robotic section had 

covered several aspects - humanoid, 

surveillance, educational and 

industrial. The robotic platform at the 

Techfest 2012 showcased bots 

expressing emotions, playing 

intelligent games and serving as a 

model for military training, among 

others. With a great potential for 

applications in future in varying 

fields, lets look at these robots one 

by one. 

Enjoy Moneycontrol.com on iPad 

and be prepared for a fantastic 

experience. Get real time stock 

quotes, interactive charts, market 

buzz, and watch CNBC-TV18, 

CNBC Awaaz live on your iPad. 

Check out the free moneycontrol app 

IIT-KGP celebrates the 9 th 

annual Alumni Meet 2012 while 

its alumni commits to give back 

to their Alma meter 

IIT Kharagpur celebrated its 9th 

Annual Alumni Meet. While it 

honoured its alumni with the 

Distinguished Service Award more 

alumni committed towards the cause 

of contributing to their Alma mater. 

A galaxy of alumni from world wide 

graced the 3-day dynamic event 

Auto Expo 2012 : Hydrogenpowered 

3-wheeler at expo 

NEW DELHI: The world's first 

hydrogen-powered three-wheeler, 

'HyAlfa', was showcased at theAuto 

Expo on Monday. Part of a 

development project dubbed 'DelHy 

3w', a fleet of 15 HyAlfa threewheelers 

will run on an experimental 

basis at Pragati Maidan, where a 


hydrogen refuelling station has also 

been set up. 

India Trade Organisation Promotion 

( ITPO) will use the vehicles on an 

experimental basis. The HyAlfa has 

been developed under a joint project 

by the United Nations Industrial 

Development Organisation ( UNIDO) 

International Centre for Hydrogen 

Energy Technologies ( ICHET), 

Mahindra & Mahindra and IIT-Delhi, 

with support from the Ministry of 

New and Renewable Energy. "The 

aim of this project is to convert 

vehicles so that they can carry and use 

hydrogen - a carbon-free fuel - and 

thus remove all pollutants," Mahindra 

& Mahindra president (automotive) 

Pawan Goenka said. He said the 

vehicle is not yet ready for 

commercial production and further 

fine-tuning will be required before 

moving in that direction. "Moreover, 

we also have to look at commercial 

viability of running a hydrogenpowered 

three-wheeler as the cost of 

hydrogen will be around Rs 250 per kg, 

which is not affordable at all," he said. 

This is a step in the right direction. 

Finally India Inc. with support from 

the likes of MNRE etc. is making 

headway in vehicles driven by 

alternative energy sources. Jury is still 

out on whether Hydrogen is the way 

of the future. Judging by the progress 

made in the USA and Germany on 

Hydrogen powered vehicles, we are 

not quite there yet. There are not only 

supply chain challenges, but 

Hydrogen continues to be a costly 

proposition. You can either produce 

Hydrogen from hydrocarbon cracking 

(which in turn has dependence on 

fossil fuels) or by splitting of water 

through electrolysis. This itself 

requires around 5-6 kWh of 

electricity. So unless the electricity 

source is a renewable source such as 

solar or wind, the electricity required 

to split water itself is most likely to 

come from thermal power plants, 

thereby not giving any benefit. 

However, as everyone knows, human 

ingenuity knows no bounds and 

technologies only develop 

incrementally. I firmly believe that we 

are one step away from a miracle 

where both Hydrogen consumption 

(by means of fuel cells etc.) and 

Hydrogen generation (by means of 

hydrocarbon cracking or 

electrolyzing/splitting water) become 

viable for mass production and 

consumption. In the short term, we can 

and must focus immediately on 

"Hydrogen supplementation". By this I 

mean - Hydrogen CNG (or HCNG in 

short). New Delhi moved to CNG 

public transport a while back. In the 

short term, this did bring down the 

pollution levels and particulate matter 

in the atmosphere. After prolonged use, 

we are now becoming aware of other 

problems such as NOX emissions due 

to unclean or inefficient burning of 

CNG. NOX is highly carcinogenic and 

now the levels of NOX in New Delhi 

are far exceeding permissible limits of 

WHO. One way to solve this problem is 

to supplement CNG with Hydrogen. A 

blended product called Hythane (trade 

name for HCNG) is already under 

experimentation by Indian Oil. In 

conclusion, while Hydrogen powered 

vehicles are all a step in the right 

direction, the government should put 

impetus on technologies that are more 

feasible in solving current big-city 

problems that Indian cities face. 

IITian tops CAT 2011 

CHENNAI: Ajinkya Deshmukh, an IIT 

- Madras graduate, decided to put in 

100% effort into preparing for the 

Common Admission Test this time, and 

got 100 percentile in return. He is one 

of nine MBA aspirants in the country to 

secure the top score this year. The IIMs 

published the CAT 2011 scores on their 

website on Wednesday. 

On his success, Ajinkya, who wrote the 

test for the third time this year, said, "I 

always thought I haven't been making a 

full effort while preparing or writing 

the test. This time I was determined to 

give my best. I told myself that I had to 

do it this year." 

Indian American sworn in as 

America's top science official 

IIT Madras alumnus, Subra Suresh, has 

been sworn in as the director of 

America's National Science Foundation 

(NSF), the top US science body with a 

$7.4 billion budget to support scientific 

institutions. 

"We are very grateful to have Subra 

taking this new task," said President 

Barack Obama at the White House 

Science after Suresh was sworn in as 

the 13th NSF director by John 

Holdren, Obama's science advisor. 

"He has been at MIT (Massachusetts 

Institute of Technology) and has 

been leading one of the top 

engineering programmes in the 

country, and for him now to be able 

to apply that to the National Science 

Foundation is just going to be 

outstanding," he said. "So we're very 

grateful for your service." 

Suresh, 54, was confirmed by the US 

Senate for a six-year term. 

He has served as dean of the 

engineering school and as Vannevar 

Bush Professor of Engineering at 

MIT. 

A mechanical engineer, who later 

became interested in materials 

science and biology, Suresh has done 

pioneering work studying the 

biomechanics of blood cells under 

the influence of diseases such as 

malaria. 

From 2000 to 2006, Suresh served as 

the head of the MIT Department of 

Materials Science and Engineering. 

He joined MIT in 1993 as the R.P. 

Simmons Professor of Materials 

Science and Engineering and held 

joint faculty appointments in the 

departments of mechanical 

engineering and biological 

engineering, as well as the division 

of health sciences and technology. 

Suresh holds a bachelor's degree 

from the Indian Institute of 

Technology in Madras and a master's 

degree from Iowa State University. 

Suresh was nominated by President 

Obama to become the new NSF 

director in place of Arden L. Bement 

Jr, who led the agency from 2004 

until he resigned in May this yea 


Dr Amitabha Ghosh was the only Asian on NASA's 

Mars Pathfinder mission. At present, he is a member of the 

Mars Odyssey Mission and the Mars Exploration Rover 

Mission. 

During the Mars Pathfinder Mission, he conducted 

chemical analysis of rocks and soil on the landing site. The 

simple and unassuming 34-year-old planetary geologist 

has won several accolades, which include the NASA Mars 

Pathfinder Achievement Award in 1997 and the NASA 

Mars Exploration Rover Achievement Award in 2004. 

The journey from India to NASA. 

It has been an intriguing experience. I was keen on 

geologic research data interpretation and solar system 

formation. During my geological research days in India, I 

had slept in railway stations while traveling to various 

places. 

After my post graduation in applied geology from IIT 

Kharagpur, I wrote a letter to a professor at NASA 

expressing a desire to work at the space agency. 

I made certain suggestions; in fact, it was a critical letter. 

In India, you can never imagine criticising your professor. 

My suggestions were approved, while I got an opportunity 

to work at NASA. 

I think one requires luck and to put in sincere effort to 

achieve one's goals. Being in the right place at the right 

time is also important. 

In Mumbai for the Pravasi Bharatiya Divas, he spoke 

about his work at NASA and his vision for India. 

The Vision for India : 

I feel there India has a great future. We have world-class 

companies. Today, companies like Infosys can be 

compared with world leaders like Oracle. Like the 

Information Technology revolution, we can have a science 

or space revolution. We have the potential to bring about 

revolutions in other sectors as well. 

Success Story 

This article contains stories/interviews of persons who succeed after graduation from different IITs 

Dr. Amitabha Ghosh 

• Post graduation in applied 

geology from IIT Kharagpur, 

• Working at NASA 

How India can we develop science and technology 

sector : 

It should be treated as a business. There should be more 

private participation. We must have an external review to 

evaluate the system and make changes as science and 

technology can take the country forward. 

We must check brain drain. About 80,000 students migrate 

to the US for further studies, and settle there. They find the 

facilities much better abroad. We need to reverse brain 

drain by enhancing and upgrading institutes in India. 

The state of space research in India : 

I don't want to make controversial statements. All I can say 

is India is not at the frontier of space research. We have 

made commendable progress but there is a long way to go. 

We can do much better. I would be glad to be of help in 

any way. Investment in research is investment in 

imagination. It is a matter of national pride and internal 

recognition. We need to allocate more funds to enhance 

research and development work. 

We need good educational institutes like IITs and IIMs, but 

IITians don't rule the world. You must remember that Microsoft 

co-founder (Bill Gates does not have a college degree. 

Youngsters must look around for role models and see what 

it is that they are doing right. Individuals must make use of 

their inherent strengths to succeed. 

How can India become a leading global player : 

Globalisation will reap huge and long-term benefits and 

India must make the best use of the opportunities. At the 

PBD seminar, I found people presenting grandiose plans. 

Instead, we should look at the realities and immediate 

solutions. 

The private sector has to be actively involved in the 

development of the country and the government has to 

respond to the needs of the people. Fifteen years ago, we 

didn't have an Infosys, today we have many global 

companies. 

There should be drastic reduction in paper work. We need 

a scenario where one can start any business in a day, like 

in the US. 

Can India have something like NASA: 

The answer is: Yes, India can. All it requires is the right 

kind of investment, infrastructure, people and support from 

the government. 


PHYSICS 

1. A large open top container of negligible mass and 

uniform cross-sectional area A has a small holes of 

cross-sectional area A/100 in its side wall near the 

bottom. The container is kept on a smooth horizontal 

floor and contains a liquid of density ρ and mass m0. 

Assuming that the liquid starts flowing out 

horizontally through the hole at t = 0, Calculate 

[IIT-1997] 

(i) the acceleration of the container, and 

(ii) its velocity when 75% of the liquid has drained 

out. 

Sol. (i) Let at any instant of time during the flow, the 

height of liquid in the container is x. The velocity 

of flow of liquid through small hole in the orifice 

by Toricelli's theorem is 

v = 2 gx 

...(i) 

The mass of liquid flowing per second through the 

orifice 

= ρ × volume of liquid flowing per second 

dm A 

= ρ × 2 gx × ...(iii) 

dt 

100 

Therefore the rate of charge of momentum of the 

system in forward direction 

dm 2gx × A × ρ 

= × v = 

(From (i) and (ii)) 

dt 100 

The rate of charge of momentum of the system in 

the backward direction 

= Force on backward direction = m × a where m is 

mass of liquid in the container at the instant t 

m = Vol. × density 

= A × x × ρ 

x 

ρ 

KNOW IIT-JEE 

v 

By Previous Exam Questions 

∴ The rate of charge of momentum of the 

system in the backward direction 

Axρ × a 

By conservation of linear momentum 

2gxAρ Axp × a = 

100 

g 

⇒ a = 

50 

(ii) By toricell's theorem v' = 2 g × ( 0. 

25h) 

where 

h is the initial height of the liquid in the container 

m0 is the initial mass 

m0 ∴ m0 = Ah × ρ ⇒ h = 

Aρ 

∴ v' = 

m 

2g 

× 0. 

25× 

Aρ 

XtraEdge for IIT-JEE 6 FEBRUARY 2012 

0 = 

gm0 2Aρ 

2. Two parallel plate capacitors A and B have the same 

separation d = 8.85 × 10 –4 m between the plates. The 

plate area of A and B are 0.04 m 2 and 0.02 m 2 

respectively. A slab of dielectric constant (relative 

permittivity) K = 9 has dimensions such that it can 

exactly fill the space between the plates of capacitor 

B. [IIT-1993] 

110 V 

(a) 

A 

(b) 

B 

A 

(c) 

(i) The dielectric slab is placed inside A as shown 

in figure (a). A is then charged to a potential 

difference of 110 V. Calculate the capacitance 

of A and the energy stored in it. 

(ii) The battery is disconnected and then the 

dielectric slab is moved from A. Find the work 

done by the external agency in removing the 

slab from A. 

(iii) The same dielectric slab is now placed inside 

B, filling it completely. The two capacitors A 

and B are then connected as shown in figure 

(c). Calculate the energy stored in the system. 

B

Sol. (i) The capacitor A with dielectric slab can be 

considered as two capacitors in parallel, one 

having dielectric slab and one not having dielectric 

A 

slab each capacitor has an area of . The 

2 

combined capacitance is 

110 V 

A 

C = C1 + C2 

( A / 2) 

ε 0 ( A / 2) 

ε r 

= + 

0ε 

= 

d 

d 

= 

0. 

4 

– 12 

× 8. 

85× 

10 

– 4 

2× 

8. 

85× 

10 

+ 

+ + + + 

– – 

A/2 

– 

A/2 

– 

+ 

A ε 0 

[1 + εr] 

2 

d 

[1 + 9] = 2 × 10 –9 F 

∴ Energy stored 

1 2 1 –9 2 –5 

= CV = × 2 × 10 × (110) = 1.21 × 10 J 

2 2 

(ii) Work done in removing the dielectric state = 

(Energy stored in capacitor without dielectric) – 

(Energy stored in capacitor with dielectric). 

It may be noted that while taking out the dielectric 

the charge on the capacitor plate remains the same. 

∴ W = 

q 

2C' 

2 

q 

– 

2C 

2 

Here, C = 2 × 10 –9 F, 

– 14 

Aε 0 0. 

04× 

8. 

85× 

10 

C' = = 

= 0.4 × 10 

– 4 

d 8. 

85× 

10 

–9 F 

q = CV = 2 × 10 –9 × 110 = 2.2 × 10 –7 C 

– 7 

2 

( 2. 

2× 

10 ) ⎡ 1 1 ⎤ 

∴ W = ⎢ – 

– 9 – 9 

2 

⎥ 

⎣0. 

4× 

10 2× 

10 ⎦ 

– 14 

2. 

2× 

2. 

2× 

10 ⎡2 

– 0. 

4⎤ 

= 

– 9 ⎢ ⎥ 

2× 

10 ⎣ 2× 

0. 

4 ⎦ 

= 1.21 × 

1. 

6 

0. 

4 

× 10 –5 = 4.84 × 10 –5 J 

ε 0ε r A B 

(iii) The capacitance of B = 

d 

– 12 

8. 

85× 

10 × 9× 

0. 

02 

= 

– 4 

8. 

85× 

10 

CB = 1.8 × 10 –9 F 

The charge on A, qA = 2.2 × 10 –7 C gets distributed 

into two parts. 

∴ q1 + q2 = 2.2 × 10 –7 C Also the potential 

difference across A = p.d. across B 


∴ 

q 1 = 

C 

A 

⇒ q1 = 

q 

C 

B 

2 

B 

C A . q2 = 

C 

0. 

4× 

10 


8× 

10 

∴ 0.22 q2 + q2 = 2.2 × 10 –7 

2. 

2 

⇒ q2 = 


22 

⇒ q1 = 0.4 × 10 –7 C 

Total energy stored 

= 

2 

1 

A 

q 

2C 

+ 

– 9 

– 9 

× 10 –7 = 1.8 × 10 –7 C 

2 

2 

q 

2C 

B 

– 14 

. q2 = 0.22 q2 

– 14 

0. 

4× 

0. 

4× 

10 1. 

8× 


8× 

10 

= 

+ 

– 9 

– 8 

2× 

0. 

4× 

10 2× 


8× 

10 

= 0.2 × 10 –5 + 0.9 × 10 –5 = 1.1 × 10 –5 J. 

3. A particles of mass m = 1.6 × 10 –27 kg and charge q = 

1.6 × 10 –19 C enters a region of uniform magnetic 

field of strength 1 Tesla along the direction shown in 

figure. The speed of the particle is 10 7 m/s. (i) the 

magnetic field is directed along the inward normal to 

the plane of the paper. The particle leaves the region 

of the field at the point F. Find the distance EF and 

the angle θ. (ii) If the direction of the field is along 

the outward normal to the plane of the paper, find the 

time spent by the particle in the region of the 

magnetic field after entering it at E. [IIT-1984] 

× × × × × 

θ × × × × × 

× × × × × 

F × × × × × 

× × × × × 

E × × × × × 

× × × × × 

× × × × × 

× × × × × 

45º 

Sol. (a) m = 1.6 × 10 –27 kg, q = 1.6 × 10 –19 C 

B = 1 T, v = 10 7 m/s 

F = q . v B sin α 

(acting towards O by Fleming's left hand rule) 

F = qvB [Q α = 90º] 

But F = ma 

∴ qvB = ma 

qvB 

∴ a = 

m 

– 19 


6× 

10 × 10 × 1 

= 

– 27 


6× 

10 

= 10 15 m/s 2 

7

O 

V 2 = 

V 

2 

θ 

F 

E 

45º 

× × × × × × × 

× × × × × × × 

× × × × × × × 

× × × × × × × 

× × × × × × × 

× × × × × × × 

× × × × × × × 

× × × × × × × 

× × 45º × × × × 

× × × × × × × 

V 

V2 = 

2 

This is the centripetal acceleration 

∴ 

v 2 

r 

= 10 15 ∠OEF = 45º 

14 

(Q OE, act as a radius) 

10 

⇒ r = = 0.1 m By symmetry 

15 

10 

∠OFE = 45º 

∴ ∠EOF = 90º (by Geometry) 

If the magnetic field is in the outward direction 

and the particle enters in the same way at E, then 

according to Fleming's Left hand rule, the particle 

will turn towards clockwise direction and cover 

3/4 th of a circle as shown in the figure. 

45º 

45º 

× × × × × × × 

E × × × × × × × 

× × × × × × × 

× × × × × × × 

× × O × × × × 

× × × × × × × 

× × × × × × × 

45º 

× × × × × × × 

× × × × × × × 

F × × × × × × × 

3 ⎡2πr ⎤ 

∴ Time required = × ⎢ ⎥ = 4.71 × 10 

4 ⎣ v ⎦ 

–8 sec. 

4. A long solenoid of radius 'a' and number of turns 

per unit length n is enclosed by cylindrical shell of 

radius R, thickness d(d

Sol. (i) Let m be the mass of electron. Then the mass 

of mu-meson is 208 m. According to Bohr's 

postualte, the angular momentum of mu-meson 

should be an integral multiple of h/2π. 

e 

r 

+3e 

nh 

∴ (208 m) vr = 

2π 

nh nh 

∴ v = 

= 

2 π× 

208mr 

416πmr 

…(1) 

Since mu-meson is moving in a circular path 

therefore it needs centripetal force which is 

provided by the electrostatic force between the 

nucleus and mu-meson. 

2 

( 208m) 

v 1 

∴ 

= 

r 4πε 

× 

3e× 

e 

2 

r 

2 

0 

3e 

∴ r = 

2 

4πε0 

× 208mv 

Substituting the value of v from (1) we get 

r = 

2 

3e 

× 416πmr 

× 416πmr 

0 

2 

4πε 

× 208n 

h 

2 2 

h ε0 

2 

n 

⇒ r = 

624πme …(2) 

(ii) The radius of the first orbit of the hydrogen 

atom 

2 

2 

ε0h 

= 

2 

πme 

…(3) 

To find the value of n for which the radius of 

the orbit is approximately the same as that of 

the first Bohr orbit for hydrogen atom, we 

equate 

equation (2) and (3) 

n 

2 2 

h ε0 

2 

624πme = 

ε 

0 

h 

2 

πme 

1 2 

(iii) = 208 R × z 

λ 

2 

⇒ n = 624 ≈ 25 

⎡ 1 1 ⎤ 

⎢ – 

2 2 ⎥ 

⎢⎣ 

n1 

n2 

⎥⎦ 

1 7 2 

⇒ = 208 × 1.097 × 10 × 3 

λ 

⇒ λ = 5.478 × 10 –11 m 

⎡ 1 1 ⎤ 

⎢ – 

2 2 ⎥ 

⎣1 

3 ⎦ 

CHEMISTRY 

6. An organic compound CxH2yOy was burnt with 

twice the amount of oxygen needed for complete 

combustion to CO2 and H2O. The hot gases, when 

cooled to 0 ºC and 1 atm pressure, measured 2.24 

L. The water collected during cooling weighed 0.9 

g. The vapour pressure of pure water at 20 ºC is 

17.5 mm Hg and is lowered by 0.104 mm Hg when 

50 g of the organic compound is dissolved in 1000 

g of water. Give the molecular formula of the 

organic compound. [IIT-1983] 

Sol. The combustion reaction is 

CxH2yOy + x O2 → x CO2 + y H2O 

To start with, the amount of O2 taken is 2x. Hence, 

after the combustion reaction, we will be left with 

the following amounts. 

Amount of oxygen left unreacted = x 

Amount of carbon dioxide = x 

Amount of water = y 

When this mixture is cooled to 0 ºC and 1 atm, we 

will be left with oxygen and carbon dioxide. Hence, 

the amount 2x occupies the given volume of 2.24 L 

at STP. Hence, 

( 2. 

24 / 2) 

L 

Amount x = 

= 0.05 mol 

− 1 

22. 

4Lmol 

Now, Mass of water collected = 0.9 g 

Amount of water collected, 

0. 

9g 

y = = 0.05 mol 

− 1 

18g 

mol 

Thus, the empirical formula of the compound is 

C0.05H2 × 0.05 O0.05, i.e. CH2O. 

Now, according to Raoult's law 

∆ p 

– = x2 

p* 


i.e. 

( 50g 

/ M) 

0. 

104 mmHg 

17. 

5mmHg 

−1 

( 50g 

/ M) 

+ ( 1000g 

/ 18g 

mol ) 

Solving for M, we get M = 150.5 g mol –1 

Number of repeating units of CH2O in the 

150. 

5 

molecular formula = 

≈ 5 

12 + 2 + 16 

Hence, Molecular formula of the compound is 

C5H10O5. 

7. The oxides of sodium and potassium contained in a 

0.5 g sample of feldspar were converted to the 

respective chlorides. The weight of the chlorides 

thus obtained was 0.1180 g. Subsequent treatment 

of the chlorides with silver nitrate gave 0.2451 g of 

silver chloride. What is the percentage of Na2O and 

K2O in the mixture ? [IIT-1979] 

=

Sol. Mass of sample of feldspar containing Na2O and 

K2O = 0.5 g. 

According to the question, 

Na2O + 2HCl → 2NaCl + H2O ..(1) 

2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g 

K2O + 2HCl → 2KCl + H2O ...(2) 

2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g 

Mass of chlorides = 0.1180 g 

Let, Mass of NaCl = x g 

∴ Mass of KCl = (0.1180 – x)g 

Again, on reaction with silver nitrate, 

NaCl + AgNO3 → AgCl + NaNO3 ...(3) 

23 + 35.5 = 58.5g 108 + 35.5 = 143.5g 

KCl + AgNO3 → AgCl + KNO3 ...(4) 

39 + 35.5 = 74.5g 108 + 35.5 = 143.5g 

Total mass of AgCl obtained = 0.2451 g 

Step 1. From eq. (3) 

58.5 g of NaCl yields = 143.5 g AgCl 

143. 

5 

∴ x g of NaCl yields = x g AgCl 

58. 

5 

And from eq. (4), 

74.5 g of KCl yields = 143.5 g of AgCl 

∴ (0.1180 – x)g of KCl yields 

143. 

5 

= (0.1180 – x)g AgCl 

74. 

5 

Total mass of AgCl 

143. 

5 

58. 

5 

x + 

143. 

5 

74. 

5 

(0.1180 – x) = 0.2451 

which gives, x = 0.0342 

Hence, Mass of NaCl = x = 0.0342 g 

And Mass of KCl = 0.1180 – 0.0342 = 0.0838g 

Step 2. From eq.(1), 

117 g of NaCl is obtained from = 62 g Na2O 

∴ 0.0342 g NaCl is obtained from 

62 

= × 0.032 = 0.018 g Na2O 

117 

From eq. (2), 

149 g of KCl is obtained from = 94 g K2O 

∴ 0.0838 g of KCl is obtained from 

94 

= × 0.0838 = 0.053 g K2O 

149 

Step 3. % of Na2O in feldspar = 

% of K2O in feldspar = 

0. 

053 

0. 

5 

0. 

018 

0. 

5 

= 3.6% 

× 100 

× 100 = 10.6 % 

8. 5 ml of 8 N of nitric acid, 4.8 ml of 5 N 

hydrochloric acid and a certain volume of 17 M 

sulphuric acid are mixed together and made upto 2 

litre. 30 ml of this mixture exactly neutralizes 42.9 

ml of sodium carbonate solution containing one 

gram of Na2CO3.10H2O in 100 ml of water. 

Calculate the amount in grams of the sulphate ions 

in solution. 

Sol. Given that, 

[IIT-1985] 

N HNO = 8 N 

3 

V HNO = 5 ml 

3 

NHCl = 5 N 

VHCl = 4.8 ml 

M H = 17 M 

2SO4 

Step 1. Meq. of HNO3 in 2L solution 

= 

N HNO × 

3 

V HNO3 

= 8 × 5 = 40 

∴ Meq. of HNO3 in 30 ml solution 

40 

= × 30 = 0.6 

2000 

Step 2. Meq. of HCl in 2L solution 

= NHCl × VHCl 

= 5 × 4.8 = 24 

∴ Meq. of HCl in 30 ml solution 

24 

= × 30 = 0.36 

2000 

Step 3. Meq. of H2SO4 in 2L solution 

= Valency factor × M × 

= 2 × 17 × 

V H2SO4 


H2SO 

4 

V H2SO4 

∴ Meq. of H2SO4 in 30 ml solution 

34× 

VH 

× 30 

2SO4 

= 

= 0.51 V H2SO4 

2000 

Step 4. Also given that, 

Volume of Na2CO3.10H2O = 100 ml 

Mass of Na2CO3.10H2O = 1 g 

Equivalent mass of Na2CO3.10H2O 

Molecular mass 286 –1 

= 

= = 143 g equiv 

2 2 

We know, 

Normality 

Mass of solute × 1000 

= 

Equivalent mass of solute × Volume (ml) 

1× 

1000 

= = 0.070 N 

143× 

100 

Meq. of Na2CO3.10H2O 

= Na2CO3. 10H2O 

V 

= 0.070 × 42.9 = 3.003 

N × Na 2CO3. 

10H2O

Step 5. 1 gram equivalent acid neutralizes 1 gram 

equivalent of base. 

∴ 0.6 + 0.36 + 0.51 V H = 3.003 

2SO4 

Solving , V H = 4 ml 

2SO4 

Step 6. 1000 ml of 1 M H2SO4 contains 

= 96 g SO4 2– ions 

96 × 17× 

4 

∴ 4ml of 17 M H2SO4 contains = 

1000 

= 6.528g SO4 2– ions 

9. The equilibrium constant Kp of the reaction 

2SO2(g) + O2(g) 2SO3(g) is 900 atm –1 at 

800 K. A mixture containing SO3 and O2 having 

initial partial pressures of 1 atm and 2 atm, 

respectively, is heated at constant volume to 

equilibrate. Calculate the pressure of each gas at 

800 K. [IIT- 1989] 

Sol. Since to start with SO2 is not present, it is expected 

that some of SO3 will decompose to give SO2 and 

O2 at equilibrium. If 2x is the partial pressure of 

SO3 that is decreased at equilibrium, we would 

have 

2SO2(g) + O2(g) 2SO3(g) 

t = 0 0 2 atm 1 atm 

teq 2x 2 atm + x 1 atm – 2x 

2 

( p ) 

2 

SO3 

( 1 atm − 2x) 

Hence, Kp = 

= 

2 

2 

( pSO 

) ( p ) 

2 O ( 2x) 

( 2 atm + x) 

2 

= 900 atm –1 

Assuming x

Sol. F is mid-point of BC i.e., F ≡ 

and AE ⊥ DE (given) 

A(i + j + k) 

λ 

B(i) 

E 

1 

F(2i) 

D 

C(3i) 

^ 

3 ^ 

i+ i 

2 

= 2 ^ i 

Let E divides AF in λ : 1. The position vector of E 

is given by 

^ 

^ 

^ 

^ 

2λ 

i+ 

1( 

i+ 

j+ 

k) 

⎛ 2λ 

+ 1⎞ 

^ 1 ^ 

= ⎜ ⎟ i + j + 

λ + 1 ⎝ λ + 1 ⎠ λ + 1 

Now, volume of the tetrahedron 

1 

= (area of the base) (height) 

3 

⇒ 

2 2 

3 

1 

= (area of the ∆ABC) (DE) 

3 

1 → → 

But area of the ∆ABC = | ( BC × BA) 

| 

2 

1 ^ ^ ^ 

1 ^ 

k 

λ + 

= | 2 i × ( j+ 

k) 

| = | i × j+ 

i× 

k) 

| 

2 

^ 

^ 

= | k – j) 

| = 2 

2 2 1 

Therefore, = ( 2) 

(DE) 

3 3 

⇒ DE = 2 

Since ∆ADE is a right angle triangle, 

AD 2 = AE 2 + DE 2 

⇒ (4) 2 = AE 2 + (2) 2 

⇒ AE 2 = 12 

But → 2λ 

+ 1 ^ 1 ^ 1 ^ ^ ^ ^ 

AE = i + j + k – ( i + j+ 

k) 

λ + 1 λ + 1 λ + 1 

^ 

= 

1 i 

λ 

^ 

– 

λ + 1 j 

λ 

^ 

– 

λ + 1 k 

λ 

λ + 

⇒ 

2 

| | 

→ 1 

AE = 

( λ + 1 

2 

2 

) 

3λ 

Therefore, 12 = 

2 

( λ + 1) 

⇒ 4(λ + 1) 2 = λ 2 

⇒ 4λ 2 + 4 + 8λ = λ 2 

⇒ 3λ 2 + 8λ + 4 = 0 

⇒ 3λ 2 + 6λ + 2λ + 4 = 0 

^ 

^ 

[λ 2 + λ 2 + λ 2 ] = 

^ 

^ 

1 

2 

3λ 

( λ + 1) 

2 

⇒ 3λ(λ + 2) + 2(λ + 2) = 0 

⇒ (3λ + 2) (λ + 2) = 0 

⇒ λ = – 2/3, λ = – 2 

Therefore, when λ = – 2/3, position vector of E is 

given by 

⎛ 2λ 

+ 1⎞ 

^ 1 ^ 1 ^ 

⎜ ⎟ i + j + k 

⎝ λ + 1 ⎠ λ + 1 λ + 1 

2.(– 

2 / 3) 

+ 1 ^ 1 ^ 1 ^ 

= 

i + j + k 

– 2 / 3 + 1 – 2 / 3 + 1 – 2 / 3 + 1 

– 4 / 3 + 1 ^ 1 ^ 1 ^ 

= i + j + k 

– 2 + 3 – 2 + 3 – 2 + 3 

3 

– 4 + 3 

3 3 

= 

3 

1/ 

3 

^ 1 ^ 1 ^ 

i + j + k 

1/ 

3 1/ 

3 

= – ^ i + 


^ 

3 j + 

^ 

3k 

and when λ = – 2 

Position vector of E is given by, 

2× 

(– 2) 

+ 1 ^ 1 ^ 1 ^ – 4 + 1 ^ 

i + j + k = i – 

– 2 + 1 – 2 + 1 – 2 + 1 – 1 

^ j – ^ k 

^ 

= 3i – ^ j – ^ k 

Therefore, – ^ ^ ^ 

i + 3 j + 3k and + ^ 

3i – ^ j – ^ k are 

the position vector of E. 

13. Evaluate ∫ π 

Sol. Let, 

−π 

/ 3 

/ 3 

3 

π + 4x 

dx [IIT-2004] 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

π / 3 πdx 

π / 3 x dx 

I = ∫ + 4 

−π 

/ 3 ⎛ π ⎞ ∫− π/ 

3 ⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

⎝ 3 ⎠ 

a ⎡ 0, 

f ( −x) 

= − f ( x) 

⎤ 

Using f ( x) 

dx= ⎢ a 

⎥ 

∫− a ⎢2 

− = ⎥ 

⎣ ∫ f ( x) 

dx, 

f ( x) 

f ( x) 

0 

⎦ 

∴ I = 2 ∫ π 

0 

/ 3 

πdx 

+ 0 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ 

⎝ 3 ⎠ 

⎧ 

⎪ 

⎨as 

⎪ 

⎪⎩ 

∫ π/ 

3 

−π 

/ 3 

π/ 

3 dx 

I = 2π∫ 0 2 − cos( x + π / 3) 

2π 

= 2π∫ dt 

2 t 

/ 3 

π / 3 − cos 

⎫ 

3 

x dx 

⎪ 

is odd ⎬ 

⎛ π ⎞ 

2 − cos⎜| 

x | + ⎟ ⎪ 

⎝ 3 ⎠ ⎪⎭ 

π 

, where x + = t 

3 

3

= 2π ∫ π 

2 / 3 

2 t 

sec dt 

2 

t 

1+ 

3tan 

2 

π/ 

3 

2 

3 2du 

= 2π∫ = 

1/ 

3 

2 

1+ 

3u 

3 

4π . { } 3 

−1 

3 

3 tan u 

4π –1 –1 4π –1 

= (tan 3 – tan 1) = tan 

3 

3 

3 

⎛ 1 ⎞ 

⎜ ⎟ 

⎝ 2 ⎠ 

π/ 

3 π + 4x 

4π –1 ⎛ 1 ⎞ 

∴ ∫ dx = tan ⎜ ⎟ . 

−π 

/ 3 ⎛ π ⎞ 

2 − cos⎜| 

x | + 

3 ⎝ 2 ⎠ 

⎟ 

⎝ 3 ⎠ 

14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 

6, is thrown n times and the list on n numbers 

showing up is noted. What is the probability that 

among the numbers 1, 2, 3, 4, 5, 6 only three 

numbers appear in this list ? [IIT-2001] 

Sol. Let us define at onto function F from A : [r1, r2 ... 

rn] to B : [1, 2, 3] where r1r2 .... rn are the readings 

of n throws and 1, 2, 3 are the numbers that appear 

in the n throws. 

Number of such functions, 

M = N – [n(1) – n(2) + n(3)] 

where N = total number of functions and 

n(t) = number of function having exactly t elements 

in the range. 

Now, N = 3 n , n(1) = 3.2 n , n(2) = 3, n(3) = 0 

⇒ M = 3 n – 3.2 n + 3 

Hence the total number of favourable cases 

= (3 n – 3.2 n + 3). 6 C3 

⇒ Required probability = 

Nuclear Physics 

1/ 

n n 6 

( 3 − 3. 

2 + 3) 

× C3 

n 

6 

3 

Physics Facts 

15. Show that the value of tanx/tan3x, wherever defined 

never lies between 1/3 and 3. [IIT-1992] 

tan x 

Sol. y = = 

tan 3x 

x – tan 


= 

= 

3tan 

tan x( 

1– 

3tan 

3tan 

1– 

3 

x – tan 

tan 

2 

2 

x 

3 – tan x 

⇒ x ≠ 0 

⇒ tan x ≠ 0 

0 ∞ 

tan x 

1– 

3tan 

2 

3 

x) 

+ – + 

1/3 3 

Let tan x = t 

⇒ y = 

3 – t 

⇒ 3y – t 2 y = 1 – 3t 2 

⇒ 3y – 1 = t 2 y – 3t 2 

2 

1– 

3t 

⇒ 3y – 1 = t 2 (y – 3) 

⇒ 

2 

3y 

– 1 

= t 

y – 3 

2 

x 

2 

3 

x 

x 

[Q tan 3x ≠ 0 ⇒ 3x ≠ 0] 

3y 

– 1 

⇒ ≥ 0, t 

y – 3 

2 ≥ 0 ∀ t ∈ R 

⇒ y ∈ (– ∞, 1/3) ∪ (3, ∞) 

Therefore, y is not defined in between (1/3, 3). 

1. Alpha particles are the same as helium nuclei and have the symbol . 

1. The atomic number is equal to the number of protons (2 for alpha) 

2. Deuterium ( ) is an isotope of hydrogen ( ) 

3. The number of nucleons is equal to protons + neutrons (4 for alpha) 

4. Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf 

generator. 

5. Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays) 

6. A loss of a beta particle results in an increase in atomic number. 

7. All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc 2 ) 

8. Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic 

numbers). 

9. Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.

1. A particle of charge Q and of negligible initial speed 

is accelerated through a potential difference of U. 

The particle reaches a region of uniform magnetic 

field of induction B, where it undergoes circular 

motion. If potential difference is doubled & B is also 

doubled then magnetic moment of the circular 

current due to circular motion of charge Q will 

become 

(A) double (B) half 

(C) four times (D) remain same 

2. Three long rod AA, AB, CC are moving with a 

speed v in a uniform magnetic field B perpendicular 

to the plane of paper as shown in figure. The 

triangle formed between the three wires is always an 

equilateral triangle. If resistance per unit length of 

wire is λ , then the induced current in the triangle is 

A B 

× × × × × × × 

B0 

× × × × × × × 

v v 

× × × × × × × 

× × × × × × × 

C 

B 

× × × × × C 

A 

(A) 

v 

3λ 

(C) 

v 

3λ 

Physics Challenging Problems 

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety 

of possible twists and turns of problems in physics, that would be very helpful in facing IIT 

JEE. Each and every problem is well thought of in order to strengthen the concepts and we 

hope that this section would prove a rich resource for practicing challenging problems and 

enhancing the preparation level of IIT JEE aspirants. 

By : Dev Sharma 

Solutions will be published in next issue 

Director Academics, Jodhpur Branch 

v B0 

B0 2B0 (B) 

λ 

3 

v 

B0 B0 (D) 

λ 

3. In the diagram shown, the wires P1Q1 and P2Q2 are 

made to slide on the rails with same speed of 5m/s. 

In this region a magnetic field of 1T exists. The 

electric current in 9 Ω resistor is 

× × P1 × × P2 × × × × 

× × × × × × × × × 

4cm × 2Ω × × × 2Ω × × × × × 

9Ω 

× × × × × × × × × 

× × × × × × × × × 

Q1 

Q2 

v 

(A) zero if both wires slide towards left 

(B) zero if both wires slide in opposite direction 

(C) 0.2mA if both wires move towards left 

(D) 0.2mA if both wires move in opposite direction 

4. A circular current loop is shown in the adjacent 

figure. The magnetic field in the region is along xaxis 

and its magnitude in the space is increasing 

with increasing y-coordinate. The net magnetic 

force on the loop is 


y 

(A) along + z-axis (B) along –z-axis 

(C) along +y-axis (D) None of these 

5. A point charge q moves from A to B along a 

parabolic path (AB is latus rectum). Electric field is 

along x-axis. The work done by the field is 

y 

B 

y 2 A B 

= 4ax 

(A) – qEa (B) –2qEa 

(C) +qeA (D) 2qEa 

6. In a cylindrical zone of radius R, magnetic field → 

B is 

present perpendicular to the plane of the paper into it. 

Magnitude of → 

B is varying with time as B0(p + qt) 

where p & q are positive constant. Consider a static 

triangle loop OAM. Emf induced in the triangular loop 

x 

x 

E 

Set # 10

(A) 

(C) 

A 

2B0πqR 

3 

B 

0 

PR 

3 

2 

× × 

× 

× 

O B × 

× 

R 

× 

× 

60º 60º 

× 

× 

× 

× 

× × × 

× × × 

× × 

× × 

2 

(B) 

B 

0 

M 

qR 

3 

2 

PB R 

(D) 

2 

2 

0 

7. Consider cylindrical region of the magnetic field 

shown in the figure. Region I and II have fields 

directed perpendicularly outward and inward 

respectively. Fields are varying with time as 

Region I : B = 3B0t 

Region II : B = B0t 

There is no net induced electric field in the region 

2 

⎛ r ⎞ 

outside the magnetic field then the ratio of 

⎜ 2 

⎟ = 

⎝ r1 

⎠ 

Region I 

× 

× 

× 

× 

× 

× 

× × 

r2 × 

× 

r1 

× 

× 

× 

× × × 

Region II 

8. Match the column: 

Column – I Column – II 

(A) In a series RLC 

circuit if C decreases 

(P) Z may increase 

(B) In a series RLC 

circuit if L increases 

(Q) Z may decrease 

(C) In a series RLC (R) resonance 

circuit if R decreases frequency increases 

(D) In a series RLC (S) power factor 

circuit if C increases decrease 

• After travelling 2.4 billion miles in just over 6 

years to reach Jupiter, Galileo missed its target at 

the Jovian moon Io by only 67 miles. That's like 

shooting an arrow from Los Angeles at a 

bull's-eye in New York and missing by only 6 

inches! 

• Utopia ia a large, smooth lying area of Mars. 

• The biggest star has a diameter of 1800 million 

miles, making it 2000 times bigger than the Sun. 

• 15% of the world's fresh water flows doen the 

Amazon. 

• In 1995, each American used an annual average 

of 731 pounds of paper, more than double the 

amount used in the 1980's. Contrary to 

predictions that computers would displace paper, 

consumption is growing. 

• The term 'black hole' was coined in 1968 when 

John Wheeler described how an in-falling object 

'becomes dimmer millisecond by 

millisecond...light and particles incident from 

outside...go down the black hole only to add to 

its mass and increase its gravitational attraction.' 

• The 'Red Planet' isn't really red at all, Nasa 

photographs indicate that it is more of a tan or 

butterscotch colour. 

• The International Space Station orbits at 248 

miles above the Earth. 

• The axis of orbit of the planet Uranus is tilted at a 

90 degree angle. 

• Astronomers have discovered over 10,000 

asteroids - but put them together and they would 

be smaller than the Moon. 

• Have you ever seen a ring around the moon? 

Folklore has it that this means bad weather is 

coming. 


1. For the process A → B 

1 3 1 

U ∝ ⇒ RT ∝ 

ρ 2 M 

V 

or T ∝ V 

So AB is isobaric process (pressure in constant) 

3. TV 

8U 

0 ⎡ 3 ⎤ 

T A = ⎢As 

4U 

0 = RTA 

3R 

⎥ 

⎣ 2 ⎦ 

M 

VA 

= 

2ρ 

0 

16U 

so 

0ρ 

0 

M 

PA 

= and VB 

= 

3M 

4ρ 

0 

so WAB = P[VB - VA] 

2. 

16U 

0ρ 

0 ⎡ M M ⎤ 16U 

0ρ 

0 M 

⇒ ⎢ − ⎥ = − × 

3M 

⎣4ρ 

0 2ρ 

0 ⎦ 3M 

4ρ0 

Option [D] is correct 

P 

The slope of the graph is 

0 

m = − 

2V0 

The equation of this graph is given by 

P0V 

3P0 

P = − + 

2V0 

2 

nRT −P0 

V 3P0 

⇒ = + 

V 2V0 

2 

−P0 

V 2 3P0 

⇒ T = − V + V 

2nRV0 

2nR 

dT −2P0V0 

3P0 

= + = 0 

dV 2nRV0 

2nR 

P0V ⇒ 

nRV0 

2P0 

3V0 

= ⇒ V = 

2nR 

2 

3V 

At 0 

V = , temp will be maximum [n = 1] 

2 

2 

P0 

9V0 

3P0 

3V0 

Tmax 

= − × + × 

2RV0 

4 2R 

2 

9 P0V0 

9 P0V0 

⇒ − + 

8 R 4 R 

9 P0V0 

⇒ 

8 R 

Option [C] is correct 

-3 PV 

= k − 3 

⇒ V = k 

nR 

for this polytrophic process, PV -2 = constant 

so x = -2 

nR( 

T2 

− T1 

) nR( 

T2 

− T1 

) 2 

w = 

= 

= nRT0 

1− 

x 

3 3 

∆U = nCV∆T = n 3/2 R(3T0 – T0) = 3nRT0 

⎛ 2 ⎞ 11 

∆ Q = ∆U 

= W = ⎜ + 3⎟nRT0 

= nRT0 

⎝ 3 ⎠ 3 

∆Q 

11 nRT 11 

∆Q = nC∆T 0 

⇒ C = = = R 

n∆T 

3 n( 

2T0 

) 6 

Option [A,C,D] is correct 

y 

4. 

R 

ωt 

F 

Fx 

Fy 

x 

F = mω 2 R 

Fy = F sin ωt 

Fy = mω 2 R sin (ωt) 

x 

* In figure as cos ωt = ⇒ x = R cos ωt 

R 

* For one complete circle, θ = ωt 

y 

* 

R 

ωt 

vx 

x vx = – ωR sin ωt 

v = ωR 

Option [A,C] is correct 

5. From 1 st 8 

Solution 

Set # 9 

Physics Challenging Problems 

Questions were Published in January Issue 

law, 

∆Q = ∆U + W 

nCαT + nCvαT + pαV 

⇒ nCαT = nCVαT + nRT/V αV 

α V 

αV 

nRT αV 

⇒ nαT0e 

αV 

= C VαT0 

e αV 

+ T0e 

αV 

V 

⇒ αC = αCV + R/V ⇒ C = CV + R/αV 

Option [B] is correct 

6. Option [C] is correct 

7. Option [B] is correct 

8. Option [D] is correct 


PHYSICS 

1. AB and CD are two ideal springs having force constant 

K1 and K2 respectively. Lower ends of these springs are 

attached to the ground so that the springs remain 

vertical. A light rod or length 3a is attached with upper 

ends B and C of springs. A particle of mass m is fixed 

with the rod at a distance a from end B and in 

equilibrium, the rod is horizontal. Calculate period of 

small vertical oscillations of the system. 

B a 

m 

2a 

C 

K1 

A D 

Sol. Let, in equilibrium, compressive forces in left and 

right springs be F1 and F2, respectively. 

Considering free body diagram of rod BC, (figure) 

B a 

2a 

C 

F1 

mg 

F1 + F2 = mg ...(i) 

Taking moments about B, 

mg × a = F2 × 3a 

1 

or F2 = mg 

3 

Substituting this value in equation (i), 

2 

F1 = mg 

3 

Let the particle be pressed from its equilibrium 

position by applying a force 'F'. Let left and right 

springs be further compressed through y1 and y2, 

respectively. Increase in compressive forces in the 

spring will be K1y1 and K2y2 respectively. In other 

words, total compressive forces in two springs will 

be (F1 + K1y1) and (F2 + K2y2) respectively. 

Now considering new free body diagram (figure) of 

the rod BC, 

Students Forum 

Expert’s Solution for Question asked by IIT-JEE Aspirants 

K2 

F2 

(F1 + K1y1) + (F2 + K2y2) = F + mg 

Taking moments about B, (figure) 

F 

...(ii) 


B 

a 

(F1 + K1y1) mg 

(F2 + K2y2) 

(F + mg) × a = (F2 + K2y2) × 3a 

Solving above equations, 

...(iii) 

or y1 = 

2a 

2 F 

F 

K1y1 = and K2y2 = . 

3 

3 

2F 

3K1 

and y2 = 

F 

3K 

2 

Since, distance of the particle from left spring is 'a' 

and that from right spring is '2a', therefore, 

downward displacement of the particle from 

equilibrium position will be 

2y1 y2 y = + = 

3 3 

4F 

9K 

l 

+ 

F 

= 

9K 

2 

C 

( K1 

+ 4K 

2 ) F 

9K 

K 

9K1K 

2 

or F = 

...(iv) 

( K1 

+ 4K 

2) 

Now, if the particle be released, it starts accelerating 

upwards due to excess compressive force in springs. 

Hence, the restoring force is (K1y1 + K2y2), which is 

numerically equal to F. 

or Restoring force = F = 

1 

9K1K 

2 . y 

( K + 4K 

) 

F 9K1K 

2 

or Restoring acceleration, a = = 

.y 

m m( 

K1 

+ 4K 

2 ) 

Since, acceleration is restoring and is directly 

proportional to displacement y, therefore, the 

particle performs SHM. Its period of oscillations is 

y 

T = 2π 

a 

2 π m( 

K1 

+ 4K 

2) 

or T = 

3 K K 

1 

2 

1 

2 

Ans. 

2

2. A plane spiral coil is made on a thin insulated wire and 

has N = 100 turns. Radii of inside and outside turns are 

a = 10 cm and b = 20 cm respectively. A magnetic field 

normal to the plane of spiral exists in the space. The 

magnetic field increases at a constant rate α = 0.3 

tesla/second. Calculate potential difference between the 

ends of the spiral. 

Sol. Since, magnetic field is increasing, therefore flux 

linked with the coil is also increasing. Due to 

increase in flux an emf is induced in it. Potential 

difference between ends of the spiral coil is equal to 

magnitude of emf induced in it. 

Since there are N turns in a radial width (b – a), 

therefore number of turns in the spiral coil per unit 

radial width is 

N 

n = 

( b – a) 

Consider a concentric circular ring of radius x and 

radial thickness dx 

Number of turns in this ring = n dx. 

Let at some instant magnetic field strength be B, 

Flux linked with the ring, φ = πx 2 B 

dφ 

Emf induced in this ring = (n . dx) 

dt 

⎛ dB ⎞ 

= (n . dx) ⎜πx 

⎟ 

⎝ dt ⎠ 

2 

∴ Total emf induced in the spiral coil, 

x= 

b 

2 

e = ∫ ( πnαx 

dx) 

= 

3 

x= 

a 

1 πn α (b 3 – a 3 ) 

= πnα x 2 dx 

Substituting value of n, 

1 2 2 

e = πN α(a + b + ab) = 2.2 volt Ans. 

3 

3. A spherical shell of radius R is filled with water. 

Temperature of atmosphere is (– θ) ºC. The shell is 

exposed to atmosphere and all water comes down to 

0ºC and then it starts freezing from outer surface 

towards the centre of the shell. Assuming shell to be 

highly conducting, calculate time for whole mass of 

water to freeze. Thermal conductivity of ice is K and 

latent heat of its fusion is L. Density of water is ρ. 

Neglect expansion during freezing. 

Sol. Heat flows from water to atmosphere because 

atmospheric temperature is below 0ºC. Water is 

filled in spherical shell, therefore heat flows in radial 

direction. 

Let at some instant t, thickness of ice layer be x. 

Then a concentric sphere of radius (R – x) is in 

liquid form as shown in figure. Heat flows from this 

sphere to atmosphere through ice layer. 

To calculate rate of heat flow, first we have to 

calculate thermal resistance of this ice layer. 

Considering a concentric spherical shell of radius r 

[(R – x) < r < R] and radial thickness dr as shown in 

figure. 


r 

(R–x) 

dr 

Its thermal resistance = 

2 

4πr 

K 

∴ Total thermal resistance of ice layer 

R 

dr 

∫ 4πKr 

( R – x) 

2 

= 

R 

x 

4πKR( 

R – x) 

Temperature just inside and outside the ice layer is 

0ºC and (– θ)ºC, respectively. 

∴ Temperature difference = 0 – (– q) = θº C 

θ 

Hence, rate of heat flow, H = 

⎛ x ⎞ 

⎜ 

⎟ 

⎝ 4πKR( 

R – x) 

⎠ 

4πKRθ( R – x) 

= 

x 

H 

rate of freezing of mass = 

L 

H 

∴ Rate of freezing of volume = 

Lρ 

But it is equal to 4π(R – x) 2 dx 

. 

dt 

∴ 

4πKRθ( 

R – x) 

Lρx 

= 4p (R – x) 2 dx 

. 

dt 

Lρx( 

R – x) 

or 

. dx = dt 

KRθ 

At instant t = 0, thickness of ice layer was equal to 

x = 0 and we have to calculate time t when whole 

mass has frozen or when x = R 

Substituting these limits, 

or t = 

t 

Lρ 

KRθ 

∫dt = ∫ 

0 

ρ 

6Kθ 

LR 2 

R 

0 

x( 

R – x) 

dx 

Ans.

4. In a Young's experiment, the upper slit is covered by a 

thin glass plate of refractive index µ1 = 1.4 while the 

lower slit is covered by another glass plate having the 

same thickness as the first one but having refractive 

index µ2 = 1.7. Interference pattern is observed using 

light of wavelength λ = 5400 Å. It is found that the 

point P on the screen where the central maxima fell 

before the glass plates ware inserted, now has 3/4 the 

original intensity. It is further observed that what used 

to be the fifth maxima earlier, lies below the point P 

while the sixth minima lies above the point P. 

Neglecting absorption of light by glass plates, calculate 

thickness of the glass plates. 

Sol. If glass plates are not inserted in slits, central 

maximum occurs at perpendicular bisector of slits. 

Hence, point P lies on perpendicular bisector of S1S2 

as shown in figure. 

S1 

S2 

Screen 

When a plate is inserted in upper slit, the 

interference pattern shifts upwards and due to plate 

inserted in lower slit, it shifts downwards. 

Since, distance of nth maxima (nth bright fringe) 

from central bright fringe is nω, where ω is fringe 

width and fifth maxima now lies below the point P, 

therefore, shift of fringe pattern is downwards and it 

is greater that 5 ω. 

Since, sixth minima (sixth dark fringe) lies above 

point P and distance of m th minima from central 

⎛ 1 ⎞ 

bright fringe is ⎜m 

– ⎟ ω, therefore, shift of fringe 

⎝ 2 ⎠ 

pattern is less than 5.5 ω. Hence, the shift lies 

between 5 ω and 5.5 ω. 

Let intensity of light due to each slit be I0. Then 

l1 = l2 = l0. 

Since, before insertion of glass plates, a bright fringe 

(central bright fringe) was formed at P, therefore, 

original intensity at P was equal to Imax. 

But Imax = ( 1 I + I 2 ) 2 = 4I0 

But now intensity at P is 3/4 of the original intensity, 

therefore now intensity at P is 

I = 3/4 Imax = 3I0. 

But I = I1 + I2 + 1 2 I I 2 cos φ where φ is phase 

difference between two rays reaching P. 

Substituting I1 = I2 = I0 and I = 3I0 in above 

equation, 

1 

cos φ = or φ = (2nπ + π/3) where n is an integer 

2 

P 

s 

But phase difference, φ = 2π where s is shift of 

ω 

fringe pattern. 

⎛ 1 ⎞ 

Hence, s = ⎜n 

+ ⎟ ω. 

⎝ 6 ⎠ 

But s lies between 5ω and 5.5 ω, therefore, 

⎛ 1 ⎞ 31 λ D 

s = ⎜5 

+ ⎟ ω = ω where ω = 

⎝ 6 ⎠ 6 

d 

31λ D 

∴ s = where D is distance of screen from 

6d 

slits and d is distance between slits. 

Let thickness of each glass plate be t. 

∴ Upward shift due to upper glass plate, 

( µ 1 – 1) 

tD 

s1 = 

d 

and downward shift due to lower glass plate, 

( µ 2 – 1) 

tD 

s2 = 

d 

∴ Resultant downward shift s = s2 – s1. 

Substituting values of s, s1 and s2, 

31λ 

t = 

6( 

µ – 

= 9.3 µm 

) 

Ans. 

2 µ 1 

5. A point isotropic light source of power P = 12 watt is 

located on the axis of a circular mirror plate of radius 

R = 3 cm. If distance of source from the plate is 

a = 39 cm and reflection coefficient of mirror plate is 

α = 0.70, calculate force exerted by light rays on the plate. 

Sol. When light rays are incident on the mirror plate, a 

part of then is reflected and a part is absorbed by the 

plate. Therefore, momentum of light rays changes. 

Due to change in the momentum, a force is 

experienced by the plate. Magnitude of the force is 

equal to rate of change of momentum. To calculate 

rate of change of momentum, first we have to 

consider a circular ring coplanar and co-axial with 

the plate. Let the radius of that ring be x and radial 

thickness dx as shown figure 

Source 


a 

Distance of every point of this ring from the source 

is 

2 2 

r = a + x . 

Hence, intensity of light incident on the ring is 

P 

I = 

2 

4πr 

θ

Direction of incident rays is inclined at angle θ with 

normal to the plane of ring. Therefore, power 

incident on the ring, p = (2πxdx)I cos θ 

P cosθx 

dx 

or p = 

2 

2r 

Since, incident power is in the form of light rays 

which are incident at angle θ with normal to the 

plate, therefore, net rate of incidence of momentum 

on the ring considered 

2 

P cos θ P cos θ x dx 

= = 

c 

2 

2r 

c 

Since, 70% of the rays are reflected and 30% are 

absorbed by the plate, therefore, rate of change of 

momentum from the ring considered 

1 

= [(0.7 p cos θ) × 2 + (0.3 p cos θ)] 

c 

But it is equal to force dF on the ring. 

Pa x dx 

∴ dF = 0.85 

2 2 2 

c( 

a + x ) 

or Total force on the plate, 

2 x= 

R 

0. 

85Pa 

x dx 

F = ∫ 2 2 

c ( a + x ) 


= 

0. 

85Pa 

2c 

2 

2 

x= 

0 

⎡ 1 

⎢ – 

2 

⎢⎣ 

a ( a 

2 

2 

2 

1 ⎤ 

2 ⎥ 

+ R ) ⎥⎦ 

0. 

85PR 

= 

= 1 × 10 

2 2 

2( 

a + R ) c 

–10 N Ans. 

NUCLEAR REACTOR TYPES – ARE THEY REALLY SAFE? 

The nuclear power industry along with the reactor technology has been constantly developing for more than five 

decades now. Nuclear reactors can be classified based on their nuclear reaction, the moderator material used, 

generation of the reactor, fuel phase, fuel type, coolant used, etc. The fission nuclear reactors are mostly dealt with 

because the fusion reactors are still in the developing stages and the fission reactors are already being used for the past 

six decades. 

• Based on nuclear reaction 

This type refers to the thermal (slow) reactors and the fast reactors based on the speed of neutrons. Thermal 

reactors are the most affordable and common as they use the natural and raw uranium; and the neutrons are 

decelerated from their natural speed when emitted from the broken atomic nuclei, and uses a moderating material 

in the process. The Fast reactors are very expensive that require more enriched fuel. 

• Based on moderator material 

Thermal reactors (because of the presence of the moderating material), and Graphite, Normal water and Heavy 

water are also used as moderators. The moderating materials in the Graphite and the Heavy water reactors 

thermalize the neutrons and keep the natural uranium intact without any enrichment. 

• Based on generation 

Generation I reactors were the first prototype reactors, Generation II used standard designs till 50s, Generation III 

were more modern, lightweighted, more efficient and were used till late 90s, the latest i.e. Generation IV reactors 

targeting on economical and minimal waste, are still in the research and development stage which may officially 

work until late 2020s. 

• Based on fuel phase and fuel type 

It is Solid, Liquid or Gas reactor where Solid is most typical. The fuel type reactors also come with fuel phaseuranium 

or thorium, which are available in abundant quantities on the land.

Matter Waves : 

Planck's quantum theory : Wave-particle duality - 

Planck gave quantum theory while explaining the 

radiation spectrum of a black body. According to 

Planck's theory, energy is always exchanged in 

integral multiples of a quanta of light or photon. 

Each photon has an energy E that depends only 

on the frequency ν of electromagnetic radiation 

and is given by : 

E = hν .....(1) 

where h = 6.6 × 10 –34 joule-sec, is Planck's 

constant. In any interaction, the photon either 

gives up all of its energy or none of it. 

From Einstein's mass-energy equivalence 

principle, we have 

E = mc 2 .....(2) 

Using equations (1) and (2), we get ; 

mc 2 = hν or m = 

hν 

.....(3) 

2 

c 

where m represents the mass of a photon in 

motion. The velocity v of a photon is equal to 

that of light, i.e., v = c. 

According to theory of relativity, the rest mass m0 

of a photon is given by : 

m0 = 

m 1− 

v 

c 

2 

2 

Here, 

hν 

m = and v = c 

2 

c 

Hence, m0 = 0 ....(4) 

i.e., rest mass of photon is zero, i.e., energy of 

photon is totally kinetic. 

The momentum p of each photon is given by : 

p = mc = 

PHYSICS FUNDAMENTAL FOR IIT-JEE 

Matter Waves, Photo-electric Effect 

hν hν 

× c = = 

2 

c c 

KEY CONCEPTS & PROBLEM SOLVING STRATEGY 

h h 

= ......(5) 

c / ν λ 

The left hand side of the above equation involves 

the particle aspect of photons (momentum) while 

the right hand side involves the wave aspect 

(wavelength) and the Planck's constant is the 

bridge between the two sides. This shows that 

electromagnetic radiation exhibits a wave- 

particle duality. In certain circumstances, it 

behaves like a wave, while in other circumstances 

it behaves like a particle. 

The wave-particle is not the sole monopoly of 

e.m. waves. Even a material particle in motion 

according to de Broglie will have a wavelength. 

The de Broglie wavelength λ of the matter waves 

is also given by : 

h h h 

λ = = = 

mv p 2mK 

where K is the kinetic energy of the particle. 

If a particle of mass m kg and charge q coulomb 

is accelerated from rest through a potential 

difference of V volt. Then 

1 2 

mv = qV or mv = 2 mqV 

2 

h 12. 

34 

Hence, λ = = Å 

2mqV 

V 

Photoelectric effect : 

When light of suitable frequency (electromagnetic 

radiation) is allowed to fall on a metal surface, 

electrons are emitted from the surface. These 

electrons are known as photoelectrons and the effect 

is known as photoelectric effect. Photoelectric 

effect, light energy is converted into electrical 

energy. 

Laws of photolectric effect : 

The kinetic energy of the emitted electron is 

independent of intensity of incident radiation. 

But the photoelectric current increases with the 

increase of intensity of incident radiation. 

The kinetic energy of the emitted electron 

depends on the frequency of the incident 

radiation. It increases with the increase of 

frequency of incident radiation. 

If the frequency of the incident radiation is less 

than a certain value, then photoelectric emission 

is not possible. This frequency is known as 

threshold frequency. This threshold frequency 

varies from emitter to emitter, i.e., depends on 

the material. 

There is no time lag between the arrival of light 

and the emission of photoelectrons, i.e., it is an 

instantaneous phenomenon. 


Failure of wave theory : 

Wave theory of light could not explain the laws of 

photoelectric effect. 

According to wave theory, the kinetic energy of 

the emitted electrons should increase with the 

increase of intensity of incident radiation. 

Kinetic energy of the emitted electron does not 

depend on the frequency of incident radiation 

according to wave theory. 

Wave theory failed to explain the existence of 

threshold frequency. 

According to wave theory there must be a time 

lag between the arrival of light and emission of 

photoelectrons. 

Einstein's theory of photoelectric effect : 

Einstein explained the laws of photoelectric effect 

on the basis of Planck's quantum theory of 

radiation. 

Einstein treated photoelectric effect as a collision 

between a photon and an atom in which photon is 

absorbed by the atom and an electron is emitted. 

According to law of conservation of energy, 

1 2 

hν = hν0 + mv 

2 

where hν is the energy of the incident photon; hv0 

is the minimum energy required to detach the 

electron from the atom (work function or 

ionisation energy) and (1/2) mv 2 is the kinetic 

energy of the emitted electron. 

The above equation is known as Einstein's 

photoelectric equation. Kinetic energy of the 

emitted electron, 

1 2 

= mv = h(ν – ν0) = hν – W 

2 

Explanation of laws of photoelectric effect : 

(a) The KE of the emitted electron increases with the 

increase of frequency of incident radiation since 

W (work function) is constant for a given emitter. 

KE is directly proportional to (ν – ν0) 

(b) Keeping the frequency of incident radiation 

constant if the intensity of incident light is 

increased, more photons collide with more atoms 

and more photoelectrons are emitted. The KE of 

the emitted electron remains constant since the 

same photon collides with the same atom (i.e., the 

nature of the collision does not change). With the 

increase in the intensity of incident light 

photoelectric current increases. 

(c) According to Einstein's equation, if the frequency 

of incident radiation is less than certain minimum 

value, the photoelectric emission is not possible. 

This frequency is known as threshold frequency. 

Hence, the frequency of incident radiation below 

which photoelectric emission is not possible is 

known as threshold frequency or cut-off 

frequency. It is given by : 

hν 

− ( 1/ 

2) 

mv 

ν0 = 

h 

On the other hand, if the wavelength of the 

incident radiation is more than certain critical 

value, then photoelectric emission is not possible. 

This wavelength is known as threshold 

wavelength of cut-off wavelength. It is given by : 

hc 

λ0 = 

2 

[ hν 

− ( 1/ 

2) 

mv ] 

(d) Since Einstein treated photoelectric effect as a 

collision between a photon and an atom, he 

explained the instantaneous nature of 


Some other important points : 

Stopping potential : The negative potential 

applied to the collector in order to prevent the 

electron from reaching the collector (i.e., to 

reduce the photoelectric current to zero) is known 

as stopping potential. 

1 2 

eV0 = mvmax. 

= hν – W = h(ν – ν0) 

2 

Millikan measured K.E. of emitted electrons or 

stopping potentials for different frequencies of 

incident radiation for a given emitter. He plotted a 

graph with the frequency on x-axis and stopping 

potential on y-axis. The graph so obtained was a 

straight line as shown in figure. 

ν0 

Frequency of incident light 


V0(stopping potential) 

2 

Millikan measured the slope of the straight line 

(=h/e) and calculated the value of Planck's constant. 

I 

Full intensity 

75% intensity 

50% intensity 

25% intensity 

V0 

– + 

Potential difference

The intercept of V0 versus ν graph on frequency 

axis is equal to threshold frequency (ν0). From 

this, the work function (hν0) can be calculated. 

Graphs in photoelectric effect : 

(a) Photoelectric current versus potential difference 

graphs for varying intensity (keeping same metal 

plate and same frequency of incident light) : 

These graphs indicate that stopping potential is 

independent of the intensity and saturation current 

is directly proportional to the intensity of light. 

ν2>ν1 

ν2 

ν1 

I 

– (V0) 2 (V0) 1 

Potential difference 

+ 

(b) Photoelectric current versus potential difference 

graphs for varying frequency (keeping same 

metal plate and same intensity of incident light) : 

These graphs indicate that the stopping potential 

is constant for a given frequency. The stopping 

potential increases with increase of frequency. 

The KE of the emitted electrons is proportional to 

the frequency of incident light. 

Stopping potential 

B1 B2 B3 ν0 

A1 A2 A3 

Frequency 

(c) Stopping potential versus frequency graphs for 

different metals : These graphs indicate that the 

stops is same for all metal, since they are parallel 

straight lines. The slope is a universal constant 

(=h/e). Further, the threshold frequency varies 

with emitter since the intercepts on frequency axis 

are different for different metals. 

Solved Examples 

1. (i) A stopping potential of 0.82 V is required to stop 

the emission of photoelectrons from the surface 

of a metal by light of wavelength 4000 Å. For 

light of wavelength 3000 Å, the stopping 

potential is 1.85 V. Find the value of Planck's 

constant. 

(ii) At stopping potential, if the wavelength of the 

incident light is kept at 4000 Å but the intensity 

of light is increased two times, will photoelectric 

current be obtained? Give reasons for your 

answer. 

Sol. (i) We have 

hc 

= eV1 + W 

λ1 


and 

⇒ 

or h = 

hc 

= eV2 + W 

λ2 

⎛ 1 1 ⎞ 

hc 

⎜ − 

⎟ = e(V2 – V1) 

⎝ λ2 λ1 

⎠ 

−19 

e( 

V2 

− V1) 


6× 

10 ( 1. 

85 − 0. 

82) 

= 

⎛ 1 1 ⎞ 8⎛ 

1 1 ⎞ 

e 

⎜ − 

⎟ 3× 

10 ⎜ − ⎟ 

⎝ λ2 

λ 

−7 

−7 

1 ⎠ ⎝ 3× 

10 4× 

10 ⎠ 

= 6.592 × 10 –34 Js 

(ii) No, because the stopping potential depends only 

on the wavelength of light and not on its intensity. 

2. A small plate of a metal (work function = 1.17 eV) is 

plated at a distance of 2m from a monochromatic 

light source of wavelength 4.8 × 10 –7 m and power 

1.0 watt. The light falls normally on the plate. Find 

the number of photons striking the metal plate per 

square metre per second. If a constant magnetic field 

of strength 10 –4 tesla is parallel to the metal surface, 

find the radius of the largest circular path followed by 

the emitted photoelectrons. 

hc 

Sol. Energy of one photon = = 

λ 

6. 

6× 

10 

−34 

4. 

8× 

10 

× 3× 

10 

−7 

= 4.125 × 10 –19 J 

Number of photons emitted per second 


0 

= 

= 2.424 × 10 

−19 

4. 

125× 

10 

18 

Number of photons striking the plate per square 

metre per second 

18 

2. 

424× 

10 

= 

= 4.82 × 10 

2 

4× 

3. 

14× 

( 2) 

16 

Maximum kinetic energy of photoelectrons emitted 

from the plate 

hc 

Emax = – W 

λ 

= 4.125 × 10 –19 – 1.17 × 1.6 × 10 –19 

= 2.253 × 10 –19 J 

8

3. A monochromatic light source of frequency 

ν illuminates a metallic surface and ejects 

photoelectrons. The photoelectrons having maximum 

energy are just able to ionize the hydrogen atom in 

ground state. When the whole experiment is repeated 

with an incident radiation of frequency (5/6) ν, the 

photoelectrons so emitted are able to excite the 

hydrogen atom beam which then emits a radiation of 

wavelength 1215 Å. Find the work function of the 

metal and the frequency ν. 

Sol. In the first case, 

Emax = Ionization energy = 13.6 eV 

= 21.76 × 10 –19 J 

So, hν = 21.76 × 10 –19 J ....(1) 

In the second case, 

So, 

hc 

E'max = 

λ 

−34 

6. 

6× 

10 × 3× 

10 

= 

−10 

1215× 

10 

=16.3×10 –19 J 

5ν h 

–19 

= 16.3 × 10 + W ...(2) 

6 

Dividing Eq.(1) by Eq.(2) 

−19 

6 21. 

76× 

10 + W 

= 

5 

−19 

16. 

3× 

10 + W 

Solving, we get 

W = 11.0 × 10 – 19 J = 6.875 eV 

−19 

−19 

21. 

76× 

10 + 11. 

0× 

10 

From Eq.(1) ν = 

−34 

6. 

6× 

10 

= 5 × 10 15 Hz 

4. The radiation, emitted when an electron jumps from 

n = 3 to n = 2 orbit in a hydrogen atom, falls on a 

metal to produce photoelectrons. The electrons from 

the metal surface with maximum kinetic energy are 

made to move perpendicular to a magnetic field of 

1/320 T in a radius of 10 –3 m. Find (i) the kinetic 

energy of electrons, (ii) wavelength of radiation and 

(iii) the work function of metal. 

Sol. (i) Speed of an electron in the magnetic field, 

Ber 

v = 

m 

Kinetic energy of electrons 

1 2 

Emax = mv = 

2 

2 

2 

2 

B e r 

2m 

2 

8 

−19 

⎛ 1 ⎞ ( 1. 

6× 

10 ) × ( 10 

= ⎜ ⎟⎠ × 

⎝ 320 

−31 

2× 

9. 

1× 

10 

= 1.374 × 10 –19 J 

= 0.8588 eV 

2 

−3 

) 

2 

(ii) Energy of the photon emitted from a hydrogen 

atom 

hc ⎡ 1 1 ⎤ 

hν = = 

λ 

⎢ − 

2 2 ⎥ 

⎣2 

3 ⎦ 

= 1.888 eV 

Wavelength of radiation, 


−34 

6. 

62× 

10 × 3× 

λ = 


888× 


6× 

10 

= 6.572 × 10 –7 m 

= 6572 Å 

8 

10 

−19 

(iii) Work function of metal W = hν – Emax 

= 1.8888 – 0.8588 

= 1.03 eV 

5. X-rays are produced in an X-ray tube by electrons 

accelerated through a potential difference of 50.0 kV. 

An electron makes three collisions in the target 

before coming to rest and loses half of its kinetic 

energy in each of the first two collisions. Determine 

the wavelengths of the resulting photons. Neglect the 

recoil of the heavy target atoms. 

Sol. Initial kinetic energy of the electron = 50.0 keV 

Kinetic energy after first collision = 25.0 keV 

Energy of the photon produced in the first collision, 

E1 = 50.0 – 25.0 = 25.0 keV 

Wavelength of this photon 

−34 

8 

hc 6. 

6× 

10 × 3× 

10 

λ1 = = 

E 

−19 

3 

1 1. 

6× 

10 × 25. 

0× 

10 

= 0.495 × 10 –10 m = 0.495 Å 

Kinetic energy of the electron after second collision 

= 12.5 eV 

Energy of the photon produced in the second 

collision, E2 = 25.0 – 12.5 = 12.5 keV 

Wavelength of this photon 

hc 

= 

E2 

6. 

6 × 10 

λ2 = 


6 × 10 

= 0.99 × 10 –10 m 

= 0.99 Å 

−34 

−19 

8 

× 3× 

10 

3 

× 12. 

5× 

10 

Kinetic energy of the electron after third collision = 0 

Energy of the photon produced in the third collision, 

E3 = 12.5 – 0 = 12.5 keV 

This is same as E2. Therefore, wavelength of this 

photon, λ3 = λ2 = 0.99 Å.

PHYSICS FUNDAMENTAL FOR IIT-JEE 

Thermal Expansion, Thermodynamics 

Thermal Expansion : 

.(a) When the temperature of a substance is increased, 

it expands. The heat energy which is supplied to 

the substance is gained by the constituent 

particles of the substance as its kinetic energy. 

Because of this the collisions between the 

constituents particles are accompanied with 

greater force which increase the distance between 

the constituent particles. 

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T 

or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ; 

V' = V(1 + γ∆T) 

(b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at 

higher temperature clearly ρ' < ρ for substances 

which have positive value of γ 

* β = 2α and γ = 3α 

Water has negative value of γ for certain temperature 

range (0º to 4ºC). This means that for that 

temperature range the volume decreases with 

increase in temperature. In other words the density 

increases with increase in temperature. 

30 ml 

25 ml 

20 ml 

15 ml 

10 ml 

5 ml 

0 ml 

If a liquid is kept in a container and the temperature 

of the system is increased then the volume of the 

liquid as well as the container increases. The 

apparent change in volume of the liquid as shown by 

the scale is 

∆Vapp = V(γ – 3α) ∆T 

Where V is the volume of liquid at lower temperature 

∆Vapp is the apparent change in volume 

γ is the coefficient of cubical expansion of liquid 

α is the coefficients of linear expansion of the 

container. 

Loss or gain in time by a pendulum clock with 

1 

change in temperature is ∆t = α(∆T) × t 

2 

KEY CONCEPTS & PROBLEM SOLVING STRATEGY 

Where ∆t is the loss or gain in time in a time interval t 

∆T is change in temperature and d is coefficient of 

linear expansion. 

If a rod is heated or cooled but not allowed to expand 

or contract then the thermal stresses developed 

F 

= γα∆T. 

A 

If a scale is calibrated at a temperature T1 but used at 

a temperature T2, then the observed reading will be 

wrong. In this case the actual reading is given by 

R = R0(1 + α∆T) 

Where R0 is the observed reading, R is the actual 

reading. 

For difference between two rods to the same at all 

temperatures l 1α1 = l2α2. 

Thermodynamics 

According to first law of thermodynamics 

q = ∆U + W 

For an isothermal process (for a gaseous system) 

(a) The pressure volume relationship is ρV = constt. 

(b) ∆U = 0 

(c) q = W 

(d) W = 2.303 nRT log10 

V f 

p 

= 2.303 nRT i log10 

Vi 

pf 

(e) Graphs T2 > T1 


P 

T2 

T1 

P 

V 

T 

T 

These lines are called isotherms (parameters at 

constant temperature) 

For an adiabatic process (for a gaseous system) 

(a) The pressure-volume relationship is PV γ = constt. 

(b) The pressure-volume-temperature relationship is 

PV 

= constt. 

T 

(c) From (a) and (b) TV γ–I = constt. 

(d) q = 0 

(e) W = –∆U 

V

(f) ∆U = ncv∆T where cv = 

piVi 

− p f V f 

(g) W = 

γ −1 

(h) Graphs 

P 

P 

= 

R 

γ −1 

nR( 

Ti 

− T f ) 

γ −1 

V 

T 

Please note that P-V graph line (isotherm) is 

steeper. 

For isochoric process 

(a) P ∝ T 

(b) W = 0 

(c) q = ∆U 

(d) ∆U = nCv∆T 

(e) Graphs 

R 

where Cv = 

γ −1 

P 

P 

V 

V 

For isobaric process 

(a) V ∝ T 

T 

(b) W = P∆V = P(Vf – Vi) = nR(Tf – Ti) 

(c) ∆U = nCv∆T 

(d) q = nCp∆T 

(e) Graphs 

P 

P 

V 

V 

T 

T 

For a cyclic process 

(a) ∆U = 0 ⇒ q = W 

(b) Work done is the area enclosed in p-V graph. 

For any process depicted by P-V diagram, area under 

the graph represents the word done. 

Kirchoff's law states that good absorbers are good 

emitters also. 

Problem solving Strategy : Thermal Expansion 

Step 1: Identify the relevant concepts: Decide 

whether the problem involves changes in length 

(linear thermal expansion) or in volume (volume 

thermal expansion) 

Step 2: Set up the problem using the following steps: 

Eq. ∆L = αL0∆T for linear expansion and 

Eq. ∆V = βV0∆T for volume expansion. 

Identify which quantities in Eq. ∆L = αL0∆T or 

∆V = βV0∆T are known and which are the 

unknown target variables. 

V 

T 

T 

Step 3: Execute the solution as follows: 

Solve for the target variables. Often you will be 

given two temperatures and asked to compute ∆T. 

Or you may be given an initial temperature T0 and 

asked to find a final temperature corresponding to 

a given length or volume change. In this case, 

plan to find ∆T first; then the final temperature is 

T0 + ∆T. 

Unit consistency is crucial, as always. L0 and ∆L 

(or V0 ∆V) must have the same units, and if you 

use a value or α or β in K –1 or (Cº) –1 , then ∆T 

must be in kelvins or Celsius degrees (Cº). But 

you can use K and Cº interchangeably. 

Step 4: Evaluate your answer: Check whether your 

results make sense. Remember that the sizes of holes 

in a material expand with temperature just as the 

same way as any other linear dimension, and the 

volume of a hole (such as the volume of a container) 

expands the same way as the corresponding solid 

shape. 

Problem solving strategy : Thermodynamics I st Law 

Step 1: Identify the relevant concepts : The first law 

of thermodynamics is the statement of the law of 

conservation of energy in its most general form. You 

can apply it to any situation in which you are 

concerned with changes in the internal energy of a 

system, with heat flow into or out of a system, and/or 

with work done by or on a system. 

Step 2: Set up the problem using the following steps 

Carefully define what the thermodynamics system is. 

The first law of thermodynamics focuses on 

systems that go through thermodynamic 

processes. Some problems involve processes 

with more than one step. so make sure that you 

identify the initial and final state for each step. 

Identify the known quantities and the target 

variables. 

Check whether you have enough equations. The 

first law, ∆U = Q – W, can be applied just once to 

each step in a thermodynamic process, so you will 

often need additional equations. These often 

include Eq. W = ∫ pdV 

for the work done in a 


V2 

V1 

volume change and the equation of state of the 

material that makes up the thermodynamic system 

(for an ideal gas, pV = nRT). 

Step 3: Execute the solution as follows : 

You shouldn't be surprised to be told that 

consistent units are essential. If p is a Pa and V in 

m 3 , then W is in joules. Otherwise, you may want 

to convert the pressure and volume units into 

units of Pa and m 3 . If a heat capacity is given in 

terms of calories, usually the simplest procedure 

is to convert it to joules. Be especially careful 

with moles. When you use n = mtot/M to convert

etween total mass and number of moles, 

remember that if mtot is in kilograms, M must be 

in kilograms per mole. The usual units for M are 

grams per mole; be careful ! 

The internal energy change ∆U in any 

thermodynamic process or series of processes in 

independent of the path, whether the substance is 

an ideal gas or not. This point is of the utmost 

importance in the problems in this topic. 

Sometimes you will be given enough information 

about one path between the given initial and final 

states to calculate ∆U for that path. Since ∆U is 

the same for every possible path between the 

same two states, you can then relate the various 

energy quantities for other paths. 

When a process consists of several distinct steps, 

it often helps to make a chart showing Q, W, and 

∆U for each step. Put these quantities for each 

step on a different line, and arrange them so the 

Q's, W's, and ∆U's form columns. Then you can 

apply the first law to each line ; in addition, you 

can add each column and apply the first law to the 

sums. Do you see why ? 

Using above steps, solve for the target variables. 

Step 4: Evaluate your answer : Check your results for 

reasonableness. In particular, make sure that each of 

your answers has the correct algebraic sign. 

Remember that a positive Q means that heat flows 

into the system, and that a negative Q means that heat 

flows into the system, and that a negative Q means 

that heat flows out of the system. A positive W 

means that work is done by the system on its 

environment, while a negative W means that work is 

done on the system by its environment. 

Solved Examples 

1. A metallic bob weighs 50 g in air. It it is immersed 

in a liquid at a temperature of 25ºC, it weighs 45 g. 

When the temperature of the liquid is raised to 100ºC, 

it weighs 45.1 g. Calculate the coefficient of cubical 

expansion of the liquid given that the coefficient of 

linear expansion of the metal is 2 × 10 –6 (ºC) –1 . 

Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm 

Weight of liquid displaced at 25ºC = V25ρ25g 

∴ 5 = V25ρ25g ...(1) 

Similarly, V100ρ100g = 50 – 45.1 = 4.9 gm ...(2) 

From eq.(1) & (2) we get, 

5 V25 

ρ25 

= . 

4. 

9 V100 

ρ100 

Now, V100 = V25(1 + γmetal × 75)= V25(1 + 3αmetal × 75) 

= V25(1 + 3 × 12 × 10 –6 × 75) 

or V100 = V25(1 + 0.0027) = V25 × 1.0027 

Also, ρ25 = ρ100(1 + γ × 75) 

where, γ = Required coefficient of expansion of the liquid 

5 

= 

4. 

9 

V ρ ( 1 + 75γ) 

1 ρ 

25 × 

V25 

× . 0027 

100 

or γ = 3.1 × 10 –4 (ºC) –1 


100 

1+ 75γ 

= 


0027 

2. A one litre flask contains some mercury. It is found 

that at different temperature the volume of air inside 

the flask remains the same. What is the volume of 

mercury in flask ? Given that the coefficient of linear 

expansion of glass = 9 × 10 –6 (ºC) –1 and coefficient of 

volume expansion of mercury = 1.8 × 10 –4 (ºC –1 ). 

Sol. Let V = Volume of the vessel 

V' = Volume of mercury 

For unoccupied volume to remain constant increase 

in volume of mercury should be equal to increase in 

volume of vessel. 

∴ V' γm∆T = Vγg∆T 

V × γ g 

or V' = 

γ m 

∴ 

−6 

1000× 

27× 

10 

V' = 

= 150 cm 

−4 


8× 

10 

3 

3. A clock with a metallic pendulum gains 6 seconds 

each day when the temperature is 20ºC and loses 12 

seconds each day when the temperature is 40ºC. Find 

the coefficient of linear expansion of the metal. 

Sol. Time taken for one oscillation of the pendulum is 

T = 2 π 

L 

g 

or T 2 = 4π 2 L 

× .....(1) 

g 

Partially differentiating, we get 

∆ L 

.....(2) 

2T∆t = 4π 2 × g 

Dividing (2) by (1), we get 

∆ T ∆ L α L∆t 

1 

= = = α∆t 

T 2L 

2L 

2 

where ∆t is the change in temperature. Now, 

One day = 24 hours = 86400 sec 

Let t be the temperature at which the clock keeps 

correct time. 

At 20ºC, the gain in time is 

1 

6 = α × (t – 20) × 86400 ....(3) 

2 

At 40ºC, the loss in time is 

1 

12 = α× (40 – t) × 86400 ...(4) 

2 

Dividing (4) by (3), we have 

12 40 − t 

= 

6 t − 20 

80 

which gives t = ºC. 

3 

Using the value in equation(3), we have 

1 ⎛ 80 ⎞ 

6 = × α × ⎜ − 20⎟⎠ 

× 86400 

2 ⎝ 3 

which gives α = 2.1 × 10 –5 perºC

4. A piston can freely move inside a horizontal cylinder 

closed from both ends. Initially, the piston separates 

the inside space of the cylinder into two equal parts 

each of volume V0, in which an ideal gas is contained 

under the same pressure p0 and at the same 

temperature. What work has to be performed in order 

to increase isothermally the volume of one part of gas 

η times compared to that of the other by slowly 

moving the piston ? 

Sol. Let volume of chambers changes by ∆V. According 

to the problem, the final volume of left chamber is η 

times final volume of right chamber. 

∴ V0 + ∆V = η(V0 – ∆V) 

or 

⎛ η −1 

⎞ 

∆V = ⎜ ⎟V0 

⎝ η + 1⎠ 

P0,v0,T0 

P0,v0,T0 

As piston is moved slowly therefore, change in 

kinetic energy is zero. By work-energy theorem, we 

can write 

ext 

Wgas in right chamber + Wgas in left chamber + W Agent = ∆KE 

ext 

W Agent = (Wgas(R) + Wgas(L)) 

We know that in isothermal process, work done is 

given by 

⎛V 

f ⎞ 

W = nRT ln ⎜ ⎟ 

⎜ ⎟ 

⎝ Vi 

⎠ 

∴ Work done by gas in left chamber (WL) 

⎛V ⎞ 

= P0V0 ln 

⎜ 0 + ∆V 

⎛ 2η 

⎞ 

⎟ = P0V0 ln ⎜ ⎟ 

⎝ V0 

⎠ ⎝ η + 1⎠ 

Similarly, work done by gas in right chamber (WR) 

⎛V ⎞ 

= P0V0 ln 

⎜ 0 − ∆V 

⎛ 2η 

⎞ 

⎟ = P0V0 ln ⎜ ⎟ 

⎝ V0 

⎠ ⎝ η + 1⎠ 

ext 

⎛ 2η 

⎞ 

W Agent = –P0V0 ln ⎜ ⎟ 

⎝ η + 1⎠ 

2 

⎛ η + 1⎞ 

= P0V0 ln ⎜ ⎟ 

⎝ 4η 

⎠ 

⎛ 2η 

⎞ 

– P0V0 ln ⎜ ⎟ 

⎝ η + 1⎠ 

5. A smooth vertical tube having two different sections 

is open from both ends equipped with two pistons of 

different areas figure. Each piston slides within a 

respective tube section. One mole of ideal gas is 

enclosed between the pistons tied with a nonstretchable 

thread. The cross-sectional area of the 

upper piston is ∆S greater than that of the lower one. 

The combined mass of the two pistons is equal to m. 

The outside air pressure is P0. By how many kelvins 

must the gas between the pistons be heated to shift 

the pistons through l. 

Sol. Let A1 = Cross section of upper piston 

A2 = Cross section of lower piston 

T = Tension in the string 

P = Gas pressure 

m1 = Mass of upper piston 

m2 = Mass of lower piston 

Now, consider FBD of upper piston 


P0 

P0 

P0 A1 

PA1 

m1g 

From equilibrium consideration of upper piston 

we get, P0A1 + T + m1g = PA1 

Similarly, consider FBD of lower piston 

T 

PA2 

P0 A2 

m2g 

∴ P0A2 + T = m2g + PA2 

Eliminating T, we get 

( m1 

+ m2) 

g 

P = P0 + 

A1 

− A2 

According to problem 

m = m1 + m2 

and ∆S = A1 – A2 

∴ 

mg 

P = P0 + 

∆ S 

Now, PV = RT 

or P∆V = R∆T 

P∆ V 

or ∆T = 

R 

But ∆V = (A1 – A2)l = ∆S. l 

⎛ mg ⎞ 

∴ ∆T = ⎜ P0 + ⎟ ∆S.l 

⎝ ∆S 

⎠ 

l 

l 

l

KEY CONCEPT 

Organic 

Chemistry 

Fundamentals 

Reduction of Aldehydes and Ketones by Hydride 

Transfer : 

R δ+ δ– 

H3B – H + C = O 

R´ 

R 

R 

– H – OH 

H – C – O H – C – O – H 

Hydride transfer Alkoxide ion Alcohol 

R 

R´ 

These steps are repeated until all hydrogen atoms 

attached to boron have been transferred. 

Sodium borohydride is a less powerful reducing 

agent than lithium aluminum hydride. Lithium 

aluminum hydride reduces acids, aldehydes, and 

ketones but sodium borohydride reduces only 

aldehydes and ketones : 

O 

Reduced by LiAlH4 

O 

C < C < C < 

O– R OR´ R R´ R 

O 

Ease of reduction 

R´ 

Reduced by NaBH4 

O 

C H 

Lithium aluminum hydride reacts violently with 

water, and therefore reductions with lithium 

aluminum hydride must be carried out in anhydrous 

solutions, usually in anhydrous ether. (Ethyl acetate 

is added cautiously after the reaction is over to 

decompose excess LiAlH4; then water is added to 

decompose the aluminum complex.) Sodium 

borohydride reductions, by contrast, can be carried 

out in water or alcohol solutions. 

The Addition of Ylides : The Wittig reaction : 

Aldehydes and ketones react with phosphorus ylides 

to yield alkenes and triphenylphosphine oxide. (An 

ylide is a neutral molecule having a negative carbon 

adjacent to a positive heteroatom.) Phosphorus ylides 

are also called phosphoranes : 

CARBONYL COMPOUND 

+ .. R´´ 

C = O + (C6H5)3P – C 

R´´´ 


R 

R 

Aldehyde or 

ketone 

Phosphorus ylide 

or phosphorane 

R 

R´ 

C = C R´´ 

+ O =P(C6H5)3 

R´´´ 

Alkene 

[(E) and(Z) isomers] 

Triphenyl phosphine 

oxide 

This reaction, known as the Wittig reaction, has 

proved to be a valuable method for synthesizing 

alkenes. The Wittig reaction is applicable to a wide 

variety of compounds, and although a mixture of (E) 

and (Z) isomers may result, the Wittig reaction offers 

a great advantage over most other alkene syntheses in 

that no ambiguity exists as to the location of the 

double bond in the product. (This is in contrast to E1 

eliminations, which may yield multiple alkene 

products by rearrangement to more stable carbocation 

intermediates, and both E1 and E2 elimination 

reactions, which may produce multiple products 

when different β hydrogens are available for 

removal.) 

Phosphorus ylides are easily prepared from 

triphenylphosphine and primary or secondary alkyl 

halides. Their preparation involves two reactions : 

General Reaction 

Reaction 1 

(C6H5)3P : + CH – X → (C6H5)3P – CH X – 

R´´ 

R´´ 

+ 

R´´´ 

R´´´ 

Triphenylphosphine An alkyltriphenylphosphonium 

halide 

Reaction 2 

(C6H5)3P – C – H : B – ⎯→ (C6H5)3P – C : – R´´ 

R´´ 

+ 

+ 

+ H:B 

R´´´ 

R´´´ 

A phosphorus ylide 

Specific Example 

Reaction 1 

(C6H5)3P : + CH3Br ⎯→ (C6H5)3P – CH3Br – 

C6H6 

+ 

Reaction 2 

Methyltriphenylphosphonium 

bromide (89%)

(C6H5)3P – CH3 + C6H5Li ⎯→ 

(C6H5)3P – CH2 : – + 

Br + 

+ C6H6 + LiBr 

– 

The first reaction is a nucleophilic substitution 

reaction. Triphenylphosphine is an excellent 

nucleophile and a weak base. It reacts readily with 1º 

and 2º alkyl halide by an SN2 mechanism to displace 

a halide ion from the alkyl halide to give an 

alkyltriphenylphosphonium salt. The second reaction 

is an acid-base reaction. A strong base (usually an 

alkyllithium or phenyllithium) removes a proton from 

the carbon that is attached to phosphorus to give the 

ylide. 

Phosphorus ylides can be represented as a hybrid of 

the two resonance structures shown here. Quantum 

mechanical calculations indicate that the contribution 

made by the first structure is relatively unimportant. 

R´´ 

(C6H5)3P = C 

(C6H5)3P – C : 

R´´´ 

–R´´ 

+ 

R´´´ 

The mechanism of the Wittig reaction has been the 

subject of considerable study. An early mechanistic 

proposal suggested that the ylide, acting as a 

carbanion, attacks the carbonyl carbon of the 

aldehyde or ketone to form an unstable intermediate 

with separated charges called a betaine. In the next 

step, the betaine is envisioned as becoming an 

unstable four-membered cyclic system called an 

oxaphosphetane, which then spontaneously loses 

triphenylphosphine oxide to become an alkene. 

However, studies by E. Vedejs and others suggest 

that the betaine is not an intermediate and that the 

oxaphosphetane is formed directly by a cycloaddition 

reaction. The driving force for the Wittig reaction is 

the formation of the very strong (∆Hº = 540 kJ mol –1 ) 

phosphorus –oxygen bond in triphenylphosphine 

oxide. 

R–C + – R ´ R ´´ R ´ R ´´ 

:C–R ´ R – C – C – R´´´ 

:O: 

Aldehyde 

or ketone 

P(C6H5)3 

+ 

.. 

– :O: 

P(C6H5)3 

+ 

Ylide Betaine 

(may not be formed) 

Specific Example : 

R´ 

R 

Alkene 

(+diastereomer) 

R ´ R ´´ 

R – C – C – R´´´ 

:O .. – P(C6H5)3 

Oxaphosphetane 

R´´ 

C = C + O = P(C6H5)3 

R´´´ 

Triphenylphosphine 

oxide 

Methylenecyclohexane 

(86%) 

– + 

O + :CH2 – P(C6H5)3 

CH2 + O=P(C6H5)3 


CH2 

O P(C6H5)3 

– 

+ 

CH2 

O –P(C6H5)3 

Michael Additions : 

Conjugate additions of enolate anions to 

α-β-unsaturated carbonyl compound are known 

generally as Michael additions. An example is the 

addition of cyclohexanone to C6H5CH=CHCOC6H5 : 

O O 

OH – 

O – 

– 

O 

C6H5CH=CH–CC6H5 

O 

O 

+H3O + 

C6H5 

CH 

CH δ– 

C—O δ– 

C6H5 

C6H5 

CH H 

C 

H 

C = O 

C6H5 

The sequence that follows illustrates how a conjugate 

aldol addition (Michael addition) followed by a 

simple aldol condensation may be used to build one 

ring onto another. This procedure is known as the 

Robinson anulation (ring-forming) reaction (after the 

English chemist, Sir Robert Robinon, who won the 

Nobel Prize in chemistry in 1947 for his research on 

naturally occurring compounds) : 

O 

O CH3 

O 

CH3 

CH2 

OH 

+ CH2 = CHCCH3 

CH2 

O 

O C 

2-Methylcyclo- 

H3C O 

hexane-1, 3-dione 

– 

CH3OH 

(conjugate 

addition) 

Methyl vinyl 

ketone 

aldol base 

condensation (–H2O) 

O 

CH3 

(65%) 

O

KEY CONCEPT 

Inorganic 

Chemistry 

Fundamentals 

Tetragonal distortion of octahedral complexes (Jahn- 

Teller distortion) : 

The shape of transition metal complexes are affected 

by whether the d orbitals are symmetrically or 

asymmetrically filled. 

Repulsion by six ligands in an octahedral complex 

splits the d orbitals on the central metal into t2g and eg 

levels. It follows that there is a corresponding 

repulsion between the d electrons and the ligands. If 

the d electrons are symmetrically arranged, they will 

repel all six ligands equally. Thus the structure will 

be a completely regular octahedron. The symmetrical 

arrangements of d electrons are shown in Table. 

Symmetrical electronic arrangements : 

Electronic 

configurat 

ion 

d 5 

d 6 

d 8 

d 10 

All other arrangements have an asymmetrical 

arrangement of d electrons. If the d electrons are 

asymmetrically arranged, they will repel some 

ligands in the complex more than others. Thus the 

structure is distorted because some ligands are 

prevented from approaching the metal. 

as closely as others. The eg orbitals point directly at 

the ligands. Thus asymmetric filling of the eg orbitals 

in some ligands being repelled more than others. This 

causes a significant distortion of the octahedral 

shape. In contrast the t2g orbitals do not point directly 

at the ligands, but point in between the ligand 

directions. Thus asymmetric filling of the t2g orbitals 

has only a very small effect on the stereochemistry. 

Distortion caused by asymmetric filling of the t2g 

orbitals is usually too small to measure. The 

electronic arrangements which will produce a large 

distortion are shown in Table. 

The two eg orbitals d 2 2 and d 2 are normally 

x − y 

z 

degenerate. However, if they are asymmetrically 

filled then this degeneracy is destroyed, and the two 

t2g 

CO-ORDINATION COMPOUND 

& METALLURGY 

eg 

orbitals are no longer equal in energy. If the d 2 

z 

orbital contains one. 

Asymmetrical electronic arrangements : 

Electronic 

configurati 

on 


d 4 

d 7 

d 9 

more electron than the d 2 2 orbital then the ligands 

x −y 

approaching along +z and –z will encounter greater 

repulsion than the other four ligands. The repulsion 

and distortion result in elongation of the octahedron 

along the z axis. This is called tetragonal distortion. 

Strictly it should be called tetragonal elongation. This 

form of distortion is commonly obsered. 

If the d 2 2 orbital contains the extra electron, then 

x −y 

elongation will occur along the x and y axes. This 

means that the ligands approach more closely along 

the z-axis. Thus there will be four long bonds and 

two short bonds. This is equivalent to compressing 

the octahedron along the z axis, and is called 

tetragonal compression, and it is not possible to 

predict which will occur. 

For example, the crystal structure of CrF2 is a 

distorted rutile (TiO2) structure. Cr 2+ is octahedrally 

surrounded by six F – , and there are four Cr–F bonds 

of length 1.98 – 2.01 Å, and two longer bonds of 

length 2.43 Å. The octahedron is said to be 

tetragonally distorted. The electronic arrangement in 

Cr 2+ is d 4 . F – is a weak field ligand, and so the t2g 

level contains three electrons and the eg level contains 

one electron. The d 2 2 orbital has four lobes whilst 

x −y 

the d 2 orbital has only two lobes pointing at the 

z 

ligands. To minimize repulsion with the ligands, the 

single eg electron will occupy the d 2 orbital. This is 

z 

equivalent to splitting the degeneracy of the eg level 

so that d 2 is of lower energy, i.e. more stable, and 

z 

d 2 2 is of higher energy, i.e. less stable. Thus the 

x −y 

t2g 

eg

two ligands approaching along the +z and –z 

directions are subjected to greater repulsion than the 

four ligands along +x, –x, +y and –y. This causes 

tetragonal distortion with four short bonds and two 

long bonds. In the same way MnF3 contains Mn 3+ 

with a d 4 configuration, and forms a tetragonally 

distorted octahedral structure. 

Many Cu(+II) salts and complexes also show 

tetragonally distorted octahedral structures. Cu 2+ has 

a d 9 configuration : 

t2g 

eg 

To minimize repulsion with the ligands, two 

electrons occupy the d 2 orbital and one electron 

z 

occupies the d 2 2 orbital. Thus the two ligands 

x − y 

along –z and –z are repelled more strongly than are 

the other four ligands. 

The examples above show that whenever the d 2 and 

z 

d 2 2 orbitals are unequally occupied, distortion 

x − y 

occurs. This is know as Jahn–Teller distortion. 

Leaching : 

It involves the treatment of the ore with a suitable 

reagents as to make it soluble while impurities 

remain insoluble. The ore is recovered from the 

solution by suitable chemical method. For example, 

bauxite ore contains ferric oxide, titanium oxide and 

silica as impurities. When the powdered ore is 

digested with an aqueous solution of sodium 

hydroxide at about 150ºC under pressure, the alumina 

(Al2O3) dissolves forming soluble sodium metaaluminate 

while ferric oxide (Fe2O3), TiO2 and silica 

remain as insoluble part. 

Al2O3 + 2NaOH → 2NaAlO2 + H2O 

Pure alumina is recovered from the filtrate 

NaAlO2 + 2H2O ⎯→ Al(OH)3 + NaOH 

2Al(OH)3 

Ignited 

⎯ 

( autoclave) 

⎯ ⎯ → Al2O3 + 3H2O 

Gold and silver are also extracted from their native 

ores by Leaching (Mac-Arthur Forrest cyanide 

process). Both silver and gold particles dissolve in 

dilute solution of sodium cyanide in presence of 

oxygen of the air forming complex cyanides. 

4Ag + 8NaCN + 2H2O + O2 

⎯→ 4NaAg(CN)2 + 4NaOH 

Sod. argentocyanide 

4Au + 8NaCN + 2H2O + O2 

⎯→ 4NaAu(CN)2 + 4NaOH 

Sod. aurocyanide 

Ag or Au is recovered from the solution by the 

addition of electropositive metal like zinc. 

2NaAg(CN)2 + Zn ⎯→ Na2Zn(CN)4 + 2Ag ↓ 

2NaAu(CN)2 + Zn ⎯→ Na2Zn(CN)4 + 2Au ↓ 

Soluble complex 

Special Methods : 

Mond's process : Nickel is purified by this method. 

Impure nickel is treated with carbon monoxide at 60– 

80º C when volatile compound, nickel carbonyl, is 

formed. Nickel carbonyl decomposes at 180ºC to 

form pure nickel and carbon monoxide which can 

again be used. 

Impure nickel + CO 60–80ºC NI(CO)4 

Gaseous compound 

Ni + 4CO 

Zone refining or Fractional crystallisation : 

Elements such as Si, Ge, Ga, etc., which are used as 

semiconductors are refined by this method. Highly 

pure metals are obtained. The method is based on the 

difference in solubility of impurities in molten and 

solid state of the metal. A movable heater is fitted 

around a rod of the impure metal. The heater is 

slowly moved across the rod. The metal melts at the 

point of heating and as the heater moves on from one 

end of the rod to the other end, the pure metal 

crystallises while the impurities pass on the adjacent 

melted zone. 

Molten zone 

containing 

impurity 

Pure metal 

Moving circular 

heater 


180ºC 

Impure 

zone 

Different metallurgical processes can be broadly 

divided into three main types. 

Pyrometallurgy : Extraction is done using heat 

energy. The metals like Cu, Fe, Zn, Pb, Sn, Ni, Cr, 

Hg, etc., which are found in nature in the form of 

oxides, carbonates, sulphides are extracted by this 

process. 

Hydrometallurgy : Extraction of metals involving 

aqueous solution is known as hydrometallurgy. 

Silver, gold, etc., are extracted by this process. 

Electrometallurgy : Extraction of highly reactive 

metals such as Na, K, Ca, Mg, Al, etc., by carrying 

electrolysis of one of the suitable compound in fused 

or molten state.

1. It is possible to supercool water without freezing. 18 

g of water are supercooled to 263.15 K(–10ºC) in a 

thermostat held at this temperature, and then 

crystallization takes place. 

Calculate ∆rG for this process. Given: 

Cp(H2O,1) = 75.312 J K –1 mol –1 

Cp (H2O,s) = 36.400 J K –1 mol –1 

∆fusH (at 0ºC) = 6.008 kJ mol –1 

Sol. The process of crystallization at 0ºC and at 101.325 

kPa pressure is an equilibrium process, for which 

∆G = 0. The crystallization of supercooled water is a 

spontaneous phase transformation, for which ∆G 

must be less than zero. Its value for this process can 

be calculated as shown below. 

The given process 

H2O(1, – 10ºC) → H2O(s, –10ºC) 

is replaced by the following reversible steps. 

(a) H2O(1, – 10ºC) → H2O(1, 0ºC) ...(1) 

273. 

15K 

∆rH1 = ∫ Cp , m ( 1) 

dT 

263. 

15K 

= (75.312 J K –1 mol –1 ) (10 K) 

= 753.12 J mol –1 

273. 

15K 

∆rS1 = ∫ 

263. 

15K 

C 

p 

, m 

R 

( 1) 

dT 

= (75.312 J K –1 mol –1 ⎛ 273. 

15K 

⎞ 

) × ln 

⎜ 

⎟ 

⎝ 263. 

15K 

⎠ 

= 2.809 J K –1 mol –1 

(b) H2O(1, 0ºC) → H2O(s, 0ºC) ...(2) 

∆rH2 = – 6.008 kJ mol –1 

– 1 

( 6008 J mol ) 

∆rS2 = – 

= – 21.995 J K 

( 273. 

15 K) 

–1 mol –1 

(c) H2O(s, 0ºC) → H2O(s, –10ºC) ...(3) 

263. 

15K 

∆rH3 = ∫ Cp , m ( s) 

dT 

273. 

15K 

= (36.400 J K –1 mol –1 )(–10 K) 

= – 364.0 J mol –1 

263. 

15K 

∆rS3 = ∫ 

273. 

15K 

C 

p, 

m 

T 

( s) 

dT 

UNDERSTANDING 

= (36.400 J K –1 mol –1 ⎛ 263. 

15K 

⎞ 

) ×ln 

⎜ 

⎟ 

⎝ 273. 

15K 

⎠ 

= – 1.358 J K –1 mol –1 

The overall process is obtained by adding Eqs. (1), 

(2) and (3), i.e. 

H2O(1, –10ºC) → H2O(s, –10ºC) 

The total changes in ∆rH and ∆rS are given by 

∆rH = ∆rH1 + ∆rH2 + ∆rH3 

=(753.12 – 6008 – 364.0) J mol –1 

= – 5618.88 J mol –1 

∆rS = ∆rS1 + ∆rS2 + ∆rS3 

= (2.809 – 21.995 – 1.358) J K –1 mol –1 

= – 20.544 J K –1 mol –1 

Now ∆rG of this process is given by 

∆rG = ∆rH – T∆rS 

= – 5618.88 J mol –1 – (263.15 K)( –20.544 J K –1 mol –1 ) 

= – 212.726 J mol –1 

2. From the standard potentials shown in the following 

diagram, calculate the potentials 

º 

E 1 and 


E1º 

º 

E 2 . 

BrO3 – 0.54 V – 

BrO 

0.45 V 1 1.07 V 

Br2 

2 

– 

Br 

0.17 V 

E2º 

Sol. The reaction corresponding to the potential Eº1 is 

BrO3 – + 3H2O + 5e – 1 

= Br2 + 6OH 

2 

– ...(1) 

This reaction can be obtained by adding the 

following two reduction reactions: 

BrO3 – + 2H2O + 4e – = BrO – + 4OH – ...(2) 

BrO – + H2O + e – 1 

= Br2 + 2OH 

2 

– 

...(3) 

Hence the free energy change of reaction (1) will be 

º 

reaction( 

1) 

Physical Chemistry 

º 

reaction( 

2) 

∆ G = ∆ G + 

∆ G 

º 

reaction( 

3) 

Replacing ∆Gºs in terms of potentials, we get 

– 5FE1º = – 4F(0.54 V) – 1F (0.45 V) 

= (–2.61 V) F

2. 

61V 

Hence E1º = = 0.52 V 

5 

Now the reaction corresponding to the potential E2º is 

BrO3 – + 2H2O + 6e – = Br – + 6OH – 

...(4) 

This reaction can be obtained by adding the 

following three reactions. 

BrO3 – + 2H2O + 4e – = BrO – + 4OH – (Eq.2) 

BrO – + H2O + e – 1 

= Br2 + 2OH 

2 

– (Eq.3) 

1 

Br2 + e 

2 

– = Br – Hence 

...(5) 

º 

reaction( 

4) 

º 

reaction( 

2) 

∆ G = ∆ G + 

∆ G 

º 

reaction( 

3) 

+ 

∆ G 

º 

reaction( 

5) 

or – 6F(E2º) = – 4F(0.54 V) – 1F(0.45 V) 

– 1F (1.07 V) 

= (– 3.68 V) F 

3. 

68 

or E2º = = 0.61 V. 

6 

3. What is the solubility of AgCl in 0.20 M NH3 ? 

Given : Ksp(AgCl) = 1.7 × 10 –10 M 2 

K1 = [Ag(NH3) + ] / [Ag + ] [NH3] = 2.33 × 10 3 M –1 and 

K2 = [Ag(NH3)2 + ]/[Ag(NH3) + ][NH3] = 7.14 × 10 3 M –1 

Sol. If x be the concentration of AgCl in the solution, then 

[Cl – ] = x 

From the Ksp for AgCl, we derive 

K sp 

−10 

[Ag + ] = 

− 

[ Cl ] 

= 


7× 

10 M 

x 

If we assume that the majority of the dissolved Ag + 

goes into solution as Ag(NH3)2 + then [Ag(NH3)2 + ] = x 

Since two molecules of NH3 are required for every 

Ag(NH3)2 + ion formed, we have [NH3] = 0.20 M – 2x 

Therefore, 

⎛ 

⎜ 

. 7× 

10 

+ 2 

[ Ag ][ NH ] 

⎜ 

3 

= 

⎝ x 

+ 

[ Ag( 

NH3 

) 2 ] 

− 

= 6.0 × 10 –8 M 2 

From which we derive 

Kinst = 

2 

2 

2 

M ⎞ 

⎟ 

( 0. 

20M 

− 2x) 

⎠ 

x 

10 

1 2 

( 0. 

20M 

− 2x) 

6. 

0× 

10 M 

= 

= 3.5 × 10 

2 

−10 

2 

x 1. 

7× 

10 M 

2 

which gives x = [Ag(NH3)2 + ] = 9.6 × 10 –3 M, which 

is the solubility of AgCl in 0.20 M NH3 

−8 

2 

4. Potassium alum is KA1(SO4)2.12H2O. As a strong 

electrolyte, it is considered to be 100% dissociated 

into K + , Al 3+ , and SO4 2– . The solution is acidic 

because of the hydrolysis of Al 3+ , but not so acidic as 

might be expected, because the SO4 2– can sponge up 

some of the H3O + by forming HSO4 – . Given a 

solution made by dissolving 11.4 g of 

KA1(SO4)2.12H2O in enough water to make 0.10 dm 3 

of solution, calculate its [H3O + ] : 

(a) Considering the hydrolysis 

Al 3+ + 2H2O Al(OH) 2+ + H3O + 

with Kh = 1.4 × 10 –5 M 

(b) Allowing also for the equilibrium 

HSO4 – + H2O H3O + + SO4 2– 

with K2 = 1.26 × 10 –2 M 


4 g 

Sol. (a) Amount of alum = 

= 0.024 mol 

− 1 

474. 

38 g mol 

0. 

024 mol 

Molarity of the prepared solution = 

3 

0. 

1dm 

= 0.24 M 

Hydrolysis of Al 3+ is 

Al 3+ + 2H2O Al(OH) 2+ + H3O + 

2+ 

+ 

[ Al( 

OH) 

][ H3O 

] 

Kh = 

3+ 

[ Al ] 

If x is the concentration of Al 3+ that has hydrolyzed, 

we have 

( x)( 

x) 

Kh = 

0. 

24M 

− x 

= 1.4 × 10–5 M 

Solving for x, we get 

[H3O + ] = x = 1.82 × 10 –3 M 

(b) We will have to consider the following equilibria. 

Al 3+ + 2H2O Al(OH) 2+ + H3O + 

H3O + + SO4 2– HSO4 – + H2O 

Let z be the concentration of SO4 2– that combines 

with H3O + and y be the net concentration of H3O + 

that is present in the solution. Since the concentration 

z of SO4 2– combines with the concentration z of 

H3O + , it is obvious that the net concentration of H3O + 

produced in the hydrolysis reaction of Al 3+ is (y + z). 

Thus, the concentration (y + z) of Al 3+ out of 0.24 M 

hydrolyzes in the solution. With these, the 

concentrations of various species in the solution are 

3+ 

Al 

0. 

24 M− 

y−z 

3 

y 

O H + + 

Thus, Kh = 

+ 2H2O 

2− 

SO4 0. 

48 M−z 

2+ 

+ 

Al( 

OH) 

+ H 


y+ 

z 

z 4 HSO − + H2O 

3 

y 

O 

( y + z)( 

y) 

= 1.4 × 10 

( 0. 

24M 

− y − z) 

–5 M ...(i)

z 

1 

K2 = 

= 

y( 

0. 

48M 

− z) 


26× 

10 

From Eq. (ii), we get 

( 0. 

48M) 

y 

z = 

2 

( 1. 

26× 

10 M) 

+ y 

− 

Substituting this in Eq. (i), we get 

−2 

M 

...(ii) 

⎛ 

⎞ 

⎜ 

( 0. 

48M) 

y 

⎟ 

⎜ 

y + 

⎟ 

y 

−2 

⎝ ( 1. 

26× 

10 M) 

+ y ⎠ 


⎛ 

⎞ 

⎜ 

( 0. 

48M) 

y 

⎟ 

⎜ 

0. 

24 − y − 

−2 

⎟ 

⎝ ( 1. 

26× 

10 M) 

+ y ⎠ 

–5 

Making an assumption that y

`tà{xÅtà|vtÄ V{tÄÄxÇzxá 

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety 

of possible twists and turns of problems in mathematics that would be very helpful in facing 

IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and 

we hope that this section would prove a rich resource for practicing challenging problems and 

enhancing the preparation level of IIT JEE aspirants. 

By : Shailendra Maheshwari 

Solutions will be published in next issue 

Joint Director Academics, Career Point, Kota 

1. Let f (x) = sin x and 

⎧{max 

f ( t); 

0 ≤ t ≤ x 

g(x) = ⎨ 2 

⎩ sin x / 2 

; 

; 

for 0 ≤ x ≤ π 

x > π 

Discuss the continuity and differentiability of g(x) in 

(0, ∞) 

2. Is the inequality sin 2 x < x sin(sin x) true for 

0 < x < π/2 ? Justify your answer. 

3. A shop sells 6 different flavours of ice-cream. In how 

many ways can a customer choose 4 ice-cream cones 

if 

(i) they are all of different flavours; 

(ii) they are not necessarily of different flavours; 

(iii) they contain only 3 different flavoures; 

(iv) they contain only 2 or 3 different flavoures ? 

4. Using vector method, show that the internal 

(external) bisector of any angle of a triangle divides 

the opposite side internally (externally) in the ratio of 

the other two sides containing the triangle. 

5. Prove that 

(a) cos x + n C1 cos 2x + n C2 cos 3x + ............ 

...... + n Cn cos(n + 1)x = 2n . cos n ⎛ n + 2 ⎞ 

x/2. cos ⎜ x⎟ 

⎝ 2 ⎠ 

(b) sin x + n C1 sin 2x + n C2 sin 3x + ............... 

....... + n Cn sin(n + 1)x = 2 n n 

. cos x/2 . sin ⎛ n + 2 ⎞ 

⎜ x⎟ 

⎝ 2 ⎠ 

6. In a town with a population of n, a person sands two 

letters to two sperate people, each of whom is asked 

to repeat the procedure. Thus, for each letter 

received, two letters are sent to separate persons 

chosen at random (irrespective of what happened in 

the past). What is the probability that in the first k 

stages, the person who started the chain will not 

receive a letter ? 

7. Prove the identity : 


x 

2 

2 

zx− 

z 

x 4 

∫ e dz = e 

0 ∫ e 

0 

x 

zx− 

z 

function f (x) = ∫ e 

0 

and solving it. 

x 2 

−z 

4 

8. Prove that ∫ sin nθsecθ dθ 

2 

dz, deriving for the 

dz a differential equation 

2cos( 

n −1) 

θ 

= – 

– 

n −1 

∫sin( 

n – 2) 

θ secθ 

dθ 

dθ. 

Hence or otherwise evaluate 

∫ π / 2 cos5θ 

sin 3θ 

dθ. 

0 cos θ 

9. Find the latus rectum of parabola 

9x 2 – 24 xy + 16y 2 – 18x – 101y + 19 = 0. 

10 

Set 

10. A circle of radius 1 unit touches positive x-axis and 

positive y-axis at A and B respectively. A variable 

line passing through origin intersects the circle in two 

points in two points D and E. Find the equation of the 

lines for which area of ∆ DEB is maximum. 

Behavior 

• Behavior is a mirror in which everyone displays 

his image. 

• Behavior is what a man does, not what he thinks, 

feels, or believes. 

• Behave the way you'd like to be and soon you'll 

be the way you behave.

MATHEMATICAL CHALLENGES 

1. as φ (a) = φ (b) = φ (c) 

so by Rolle’s theorem there must exist at least a point 

x = α & x = β each of intervals (a, c) & (c, b) such 

that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem, 

there must exist at least a point x = µ such that 

α < µ < β where φ′(µ) = 0 

2 f ( a) 

2 f ( b) 

so 

+ 

( a − b) 

( a − c) 

( b − c) 

( b − a) 

2 f ( c) 

+ 

– f ′′ (µ) = 0 

( c − a) 

( c − b) 

f ( a) 

f ( b) 

so 

+ 

( a − b) 

( a − c) 

( b − c) 

( b − a) 

f ( c) 

1 

+ 

= f ′′ (µ) 

( c − a) 

( c − b) 

2 

where a < µ < b. 

2. Required probability 

r 

5 5 5 5 1 5 

1 . . . ........ . = 

6 6 6 6 6 6 

− 

⎛ ⎞ 

⎜ ⎟ 

⎝ ⎠ 

2 

1 

. (r – 2) times 

6 

Note : any number in 1st loss 

same no. does not in 2nd (any other comes). 

Now 3rd is also diff. (and in same r − 2 times) 

Now (r − 1) th & r th must be same. 

3. 2s = a + b + c 

ON = − BN + BO 

Let BN = x 

2BN + 2CN + 2AR = 2s 

x + (a − x) + (b − a + x) = s 

x = s − b 

A 

B 

M 

I (h,k) 

N 

r 

a 

so h = ON = − (s − b) 

2 

−2s 

+ a + 2b 

b − c 

= 

= & r = k. 

2 2 

SOLUTION FOR JANUARY ISSUE (SET # 9) 

R 

O 

C 

∆ 

so r = k = = 

s 

s ( s − a)( 

s − b)( 

s − c) 

s 

r = k = 

s ( s − a)( 

s − b)( 

s − c) 

s 

2sk = s( s − a)( 

a − b + c)( 

a + b − c) 

= s ( s − a)( 

a − 2x)( 

a + 2x) 

2sk = s( s − a)( 

a − 4h 

) 

required locus is 

4s 2 y 2 = A(a 2 – 4x 2 ) 

⇒ s2y2 + Ax2 Aa 

= 

4 

where A is = s (s – a) 

here h 2 < as so it is an ellipse 

4. f (0) = c 

f (1) = a + b + c & f (−1) = a − b + c 

solving these, 

1 

a = [f (1) + f (−1) − 2 f (0)] , 

2 

1 

b = [f (1) − f (−1)] & c = f (0) 

2 

so f (x) = 

x( 

x + 1) 


2 

2 

2 

2 

f (1) + (1− x 2 ) f (0) + 

x( 

x −1) 

2 

f(−1) 

2 | f (x) | < | x | | x + 1 | + 2| 1 − x 2 | + | x | | x − 1| ; 

as | f (1) | , | f (0) |, | f (−1) | ≤ 1. 

2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x 2 ) + | x | (1 − x) as 

x ∈ [−1, 1] 

so 2 | f (x) | ≤ 2 (|x| + 1 − x 2 5 5 

) ≤ 2 . so | f (x) | ≤ 

4 

4 

Now as g (x) = x 2 1 

f (1/x) = (1 + x) f (1) 

2 

+ (x 2 1 

− 1) f (0) + (1 − x) f (−1) 

2 

so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x 2 | + | 1 − x| 

⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x 2 ) | + 1 − x ; 

as x ∈ [−1, 1] 

⇒ 2 | g (x) | < − 2x 2 + 4 ≤ 4. 

⇒ |g (x) | ≤ 2. 

5. Oil bed is being shown by the plane A′ PQ. θ be the 

angle between the planes A′ PQ & A′ B′ C′. Let A′ B′ 

C′ be the x − y plane with x-axis along A′ C′ and 

origin at A′. The P.V.s of the various points are 

defined as follows

B 

A C 

A´ 

B´ 

P 

point C′ : b î , point B′ : cos A î + c sin A j ˆ , 

point Q : b î – z kˆ , point P : cos A î + c sin A j ˆ – y kˆ normal vector to the plane A′ B′ C′ 

C´ 

= n1 r = bc sin A kˆ normal vector to the plane A'PQ = n2 r 

= cz sin A î + (by – cz cos A) jˆ + bc sin A kˆ r r 

n1. n2 

so cos θ = r r 

| n || n | 

1 

1 

bc sin A 

= 

2 2 2 

2 2 2 2 

[ c z sin A + ( by − cz cos A) 

+ b c sin A] 

b c sin A 

cos θ = 

2 2 2 2 2 2 2 

[ b c sin A + ( c z + b y − 2bycz 

cos A)] 

2 

[ c z 

so tan θ = 

2 

2 

2 

Q 

+ b y − 2bycz 

cos A] 

bc sin A 

1/ 

2 

z y 2yz 

so tan θ . sin A = + − cos A 

2 2 

b c bc 

6. 

cos8x 

− cos 7x 

2sin 

5x 

∫ . dx 

1+ 

2cos5x 

2sin 

5x 

= ∫ 

sin13x 

−sin 

3x 

−sin12x 

+ sin 2x 

dx 

2 (sin 5x 

+ sin10x) 

= ∫ 

sin13x 

+ sin 2x 

−sin 

3x 

– sin12x 

dx 

2 (sin 5x 

+ sin10x) 

2 

= ∫ 

15x 

11x 

15x 

9x 

2sin 

cos − 2sin 

cos 

2 2 2 2 dx 

15x 

5x 

2. 

2. 

sin cos 

2 2 

= ∫ 

11x 

9x 

cos − cos 

2 2 dx 

5x 

2 cos 

2 

x 

− 2sin 

5x 

sin 

= 2 

∫ dx 

5x 

2cos 

2 

2 

1/ 

2 

1/ 

2 

= − 2 ∫ 

⎛ 5x 

⎜sin 

sin 

⎝ 2 

x ⎞ 

⎟ dx 

2 ⎠ 

= ∫ 

⎛ 6x 

4x 

⎞ 

⎜cos 

− cos ⎟ dx 

⎝ 2 2 ⎠ 

= ∫ (cos 3x 

− cos 2x) 

dx 


7. 

= 

sin 3x 

− 

3 

sin 2x 

+ C 

2 

2 x 

d y 

= 2 

2 

dx ∫ f ( t) 

dt 

0 

integrate using by parts method 

dy 

⎡ x 

x ⎤ 

= 2 ⎢x 

− ⎥ 

dx ⎢ ∫ f ( t) 

dt ∫ x . f ( x) 

dx 

⎥ 

⎣ 0 

0 ⎦ 

⎡ x 

⎤ 

= 2 ⎢ 

⎥ 

⎢∫ 

( x − t) 

f ( t) 

dt 

⎥ 

⎣ 0 

⎦ 

again integrating, 

⎡ x 

x ⎛ x ⎞ ⎤ 

y = 2 ⎢ x 

⎜ 

⎟ ⎥ 

⎢ ∫( 

x − t) 

f ( t) 

dt − ∫ x − 

⎜∫ 

f ( t) 

dt 0 dx 

⎟ ⎥ 

⎣ 0 

0 ⎝ 0 ⎠ ⎦ 

⎡ 

=2 ⎢x 

⎢ 

⎣ 

x 

x 

∫ 

0 

2 x 

x 

x 

( x − t) 

f ( t) 

dt − ∫ f ( t) 

dt + 

2 ∫ 2 

2 

= ∫ 2 ( x − xt) 

f ( t) 

dt − ∫ x 

0 

x 

x 

0 

0 

2 

x 

0 

x 

2 

f ( t) 

dt + ∫ t 

y = ∫ ( x − 2xt 

+ t ) f ( t) 

dt = ∫( x − t) 

0 

2 

2 

1/ 

α 

8. 

⎛ α ⎞ 

To prove that ⎜⎛ 

a ⎞ 

⎜ ⎟ + 1⎟ 

⎜ ⎟ 

⎝⎝ 

b ⎠ ⎠ 

a 

Let = c > 0 

b 

so (c 

⎛ β ⎞ 

< ⎜⎛ 

a ⎞ 

⎜ ⎟ + 1⎟ 

⎜ ⎟ 

⎝⎝ 

b ⎠ ⎠ 

α + 1) 1/α < (cβ + 1) 1/β . 

Let f (x) = (cx + 1) 1/x ; x > 0 

f ′(x) = (cx + 1) 1/x ln (cx ⎛ 1 ⎞ 

+ 1) ⎜− 

2 ⎟ 

⎝ x ⎠ 

x 

1 

−1 

x 

x 

0 

2 

0 

⎤ 

f ( x) 

dx⎥ 

⎥ 

⎦ 

2 

f ( t) 

dt 

1/ 

β 

1 1 

+ (cx + 1) 

–1. 

x cx ln c 

x 

f ( t) 

dt 

( c + 1) 

x 

x x x 

= 

[ − ( c + 1) 

l n ( c + 1) 

+ c ln 

c ] < 0 

2 

x 

so f (x) is decreasing function 

so f (α) < f (β). Hence proved.

9. Point P (x, 1/2) under the given condition are length 

PB = OB 

O 

(t – 1) 

P 

A 

θ 

C 

B (t, 1) 

rθ = t ; so θ = t 

PB θ 

from ∆PAB : = PA sin 

2 2 

t 

⇒ PB = 2 sin ........(1) 

2 

θ t 

Now ∠ PBC = = ; 

2 2 

θ t 

so from ∠ PCB ; = 

2 2 

1/ 

2 t 

so from ∆ PCB ; = sin 

PB 2 

........(2) 

from (1) & (2) PB = 1 ; so θ = t = π/3 

thus | PB | 2 = (t − x) 2 1 

+ = 1. 

4 

3 3 

| t − x | = ; t − x = ; as t > x 

2 

2 

π 3 

so x = − 

3 2 

10. Let xn = n −1 

+ n + 1 be rational, then 

1 

= 

xn 

1 

n −1 + 

is also rational 

n + 1 

1 

= 

xn 

n + 1 − 

2 

n −1 


n + 1 − n −1 


as n + 1 + n −1 

& n + 1 − n −1 

are rational 

so n + 1 + n −1 

must be rational 

i.e. (n + 1) & (n – 1) are perfect squares. 

This is not possible as any two perfect squares differe 

at least by 3. Hence there is not positive integer n for 

which n −1 

+ n + 1 is a rational. 

Regents Physics 

Modern Physics : 

You Should Know 

• The particle behavior of light is proven by the 


• A photon is a particle of light {wave packet}. 

• Large objects have very short wavelengths when 

moving and thus can not be observed behaving 

as a wave. (DeBroglie Waves) 

• All electromagnetic waves originate from 

accelerating charged particles. 

• The frequency of a light wave determines its 

energy (E = hf). 

• The lowest energy state of a atom is called the 

ground state. 

• Increasing light frequency increases the kinetic 

energy of the emitted photo-electrons. 

• As the threshold frequency increase for a photocell 

(photo emissive material) the work function 

also increases. 

• Increasing light intensity increases the number of 

emitted photo-electrons but not their KE. 

Mechanics : 

• Centripetal force and centripetal acceleration 

vectors are toward the center of the circle- while 

the velocity vector is tangent to the circle. 

• An unbalanced force (object not in equilibrium) 

must produce acceleration. 

• The slope of the distance-tine graph is velocity. 

• The equilibrant force is equal in magnitude but 

opposite in direction to the resultant vector. 

• Momentum is conserved in all collision systems. 

• Magnitude is a term use to state how large a 

vector quantity is. 


1. If 

MATHS 

4 

sin α 

8 

a 

+ 

4 

cos α 

8 

b 

= 

1 

a + b 

sin α cos α 1 

+ = 

3 

3 

a b ( a + b) 

3 

, show that 

a + b 4 a + b 4 

Sol. Here sin α + cos α = 1 

a 

b 

or sin 4 α + cos 4 b 4 a 4 

α + sin α + cos α = 1 

a b 

or (sin 2 α + cos 2 α) 2 – 2 sin 2 α . cos 2 α 

b 4 a 4 

+ sin α + cos α = 1 

a b 

or 

or 

∴ 

⎛ 

⎜ 

⎝ 

⎛ 

⎜ 

⎝ 

b 

a 

b 

a 

2 

sin 

sin 

2 

⎞ 

α⎟ 

⎟ 

⎠ 

2 

α – 

. 

b 2 a 2 

– 2 . sin α . cos α 

a b 

a 

b 

cos 

b 2 a 2 

sin α = cos α 

a b 

2 

⎞ 

α⎟ 

⎟ 

⎠ 

2 

⎛ 

+ ⎜ 

⎝ 

= 0 

or sin 2 a 2 

α = cos α 

b 

2 

2 

2 

sin α cos α sin α + cos 

∴ = = 

a b a + b 

∴ sin 2 a 

α = 

a + b 

, cos2 b 

α = 

a + b 

8 

sin α cos α 

∴ + = 

3 

3 

a b 

= 

a 

( a + b 

4 

) 

+ 

8 

b 

( a + b 

1 

3 

a . 

4 

) 

a 

4 

( a + b) 

= 

4 

a + b 

( a + b 

2 

4 

) 

+ 

a 

b 

α 

2 

cos 

1 

3 

b . 

= 

2 

⎞ 

α⎟ 

= 0 

⎟ 

⎠ 

b 

4 

( a + b) 

1 

( a + b) 

2. Let [x] stands for the greatest integer function find 

2 

3 x + sin x 

the derivative of f(x) = ( x + [ x + 1]) 

, where it 

exists in (1, 1.5). Indicate the point(s) where it does 

not exist. Give reason(s) for your conclusion. 

Students' Forum 

Expert’s Solution for Question asked by IIT-JEE Aspirants 

3 

4 

Sol. The greatest integer [x 3 + 1] takes jump from 2 to 3 at 

3 3 

2 and again from 3 to 4 at 3 in [1, 1.5] and 

therefore it is discontinuous at these two points. As a 

result the given function is discontinuous at 3 2 and 

hence not differentiable. 

To find the derivative at other points we write : 

in (1, 3 2 ), f(x) = ( x + 2) 

⇒ f ´(x) = ( x + 2) 

2 

x + sin x−1 

2 

x + sin x 

{x 2 + sin x + (x + 2) (2x + cos x) log (x + 2)} 

2 

x + sin x 

in ( 3 2, 3 3 ), f(x) = ( x + 3) 

, 

f ´(x) = ( 

2 

x + sin x−1 

x + 3) 

{x 2 + sin x 

+ (2x + cos x) (x + 3) × loge (x + 3)} 

2 

x + sin x 

in ( 3 5 , 1.5), f(x) = ( x + 4) 

, 

2 

x + sin x−1 

x , {x 2 + sin x + (2x + cos x) 

f ´(x) = ( + 4) 

(x + 4) × loge(x + 4)} 

3. The decimal parts of the logarithms of two numbers 

taken at random are found to six places of decimal. 

What is the chance that the second can be subtracted 

from the first without "borrowing"? 

Sol. For each column of the two numbers, 

n(S) = number of ways to fill the two places by the 

digits 0, 1, 2, ... , 9 

= 10 × 10 = 100. 

x 

× × × × × × 

y 

× × × × × × 

Let E be the event of subtracting in a column without 

borrowing. If the pair of digits be (x, y) in the column 

where x is in the first number and y is in the second 

number then 

E = {(0, 0), (1, 0), (2, 0), .. ,(9, 0), 

(1, 1), (2, 1), ..., (9, 1), 

(2, 2), (3, 2), ..., (9, 2), 

(3, 3), (4, 3), ..., (9, 3), 

...... 

(8, 8), (9, 8), 

(9, 9)} 


10. 

11 

∴ n(E) = 10 + 9 + 8 + ... + 2 + 1 = = 55 

2 

∴ the probability of subtracting without borrowing 

55 

in each column = . 

100 

⎛ 55 ⎞ ⎛ 11 ⎞ 

∴ the required probability = ⎜ ⎟⎠ = ⎜ ⎟⎠ . 

⎝100 

⎝ 20 

4. Let S be the coefficients of x 49 in given expression 

f(x) and if P be product of roots of the equation 

S 

f(x) = 0, then find the value of , given that : 

P 

f(x) = (x – 1) 2 ⎛ x ⎞ ⎛ 1 ⎞ ⎛ x ⎞ ⎛ 1 ⎞ 

⎜ − 2⎟ 

⎜ x − ⎟ ⎜ − 3⎟ 

⎜ x − ⎟ , 

⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ 

⎛ x ⎞ ⎛ 1 ⎞ 

......... ⎜ − 25⎟ 

⎜ x − ⎟ 

⎝ 25 ⎠ ⎝ 25 ⎠ 

Sol. Here we can write f(x) as : 

⎧ ⎛ x ⎞⎛ 

x ⎞ ⎛ x ⎞⎫ 

f(x) = ⎨( 

x −1) ⎜ − 2⎟⎜ 

− 3⎟... 

⎜ − 25⎟⎬ 

⎩ ⎝ 2 ⎠⎝ 

3 ⎠ ⎝ 25 ⎠⎭ 

⎧ ⎛ 1 ⎞⎛ 

1 ⎞ ⎛ 

× ⎨( 

x −1) 

⎜ x − ⎟⎜ 

x − ⎟... 

⎜ x − 

⎩ ⎝ 2 ⎠⎝ 

3 ⎠ ⎝ 

Now roots of f(x) = 0 are; 

1 2 , 2 2 , 3 2 , ..... , 25 2 1 1 1 

and 1, , , ....., 

2 3 25 

Now f(x) is the polynomial of degree 50, 

So coefficient of x 49 will be : 

S = – (sum of roots) 

= – (1 2 + 2 2 + ... + 25 2 ⎛ 1 1 1 ⎞ 

) – ⎜1 

+ + + .... + ⎟ 

⎝ 2 3 25 ⎠ 

6 

25 

6 

1 ⎞⎫ 

⎟⎬ 

25 ⎠⎭ 

⎧25× 

26× 

51 ⎫ 

1 

= – ⎨ + K⎬ 

where, K = 

⎩ 6 

∑ ⎭ 

n 

n= 

1 

⇒ S = –(K + 5525). 

Product of roots : 

1 2 . 2 2 . 3 2 .... 25 2 1 1 1 

. 1 . . .... = 1 . 2 . 3 ...25 

2 3 25 

∴ P = 25 ! 

25 

S −( 

K + 5525) 

1 

Hence = 

, where K = 

P 25! 

∑ n 

n= 

1 

5. A traveller starts from a certain place on a certain day 

and travels 1 km on the first day and on subsequent 

days, he travels 2 km more than the previous day. 

After 3 days, a second traveller sets out from the 

same place and on his first day he travels 12 km and 

on subsequent days he travels 1 km more than the 

previous day. On how many days will the second 

traveller be ahead of the first? 

Sol. The first traveller travels on different days as follow 

(in km) 1, 1 + 2, 1 + 2 + 2, ... . 

After 3 days the first traveller is already ahead by 

(1 + 3 + 5) km, i.e., 9 km. 

1 3 5 

The second traveller travels on different days as 

follows : 0, 0, 0, 12, 13, 14, ... 

After n days from the day the second traveller starts, 

the distance covered by the first 

= 1 + 3 + 5 + (7 + 9 + ... to n terms) 

= 1 + 3 + 5 + ... to (n + 3) terms 

= (n + 3) 2 

and the distance covered by the second 

= 12 + 13 + 14 + ... to n terms 

n n( 

n + 23) 

= {24 + (n – 1).1} = 

2 

2 

The second traveller is ahead of the first on the n th 

day (after the second sets off) if 

n( 

n + 23) 

> (n + 3) 

2 

2 

or n 2 + 23n > 2(n 2 + 6n + 9) 

or n 2 – 11n + 18 < 0 

or (n – 2) (n – 9) < 0. 

So n – 2 > 0 and n – 9 < 0 ...(i) 

or n – 2 < 0 and n – 9 > 0 ...(ii) 

(i) ⇒ n > 2 and n < 9 

(ii) ⇒ n < 2 and n > 0 (absurd) 

Thus, from the begining of the 3 rd day to the end of 

the 9 th day the second traveller is ahead of the first. 

So, the second is ahead of the first on the 3 rd , 4 th , 5 th , 

..., 9 th days (after the second sets off). 

Hence, the required number of days = 7. 

6. Let P(x) be a polynomial of degree n such that 

i 

P(i) = for i = 0, 1, 2 ..... n. If n is odd than find 

i + 1 

the value of P(n + 1). 

Sol. Let Q(x) = (x + 1) P(x) – x 

clearly Q(x) is polynomial of degree n + 1. Also 

i 

Q(i) = (i + 1) – i = 0 for i = 1, 2, 3 .....n 

i + 1 

Thus we can assume 

Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a constant. 

Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n) 

1 = (–1) n + 1 k(n + 1) ! 

⇒ k = 


( n 

1 

+ 

1) 

! 

(Q n is odd) 

Thus P(x) = 

1 ⎡ x( 

x −1)( 

x − 2)....( 

x − n) 

⎤ 

⎢ 

+ x⎥ 

, 

x + 1 ⎣ ( x + 1) 

! ⎦ 

where n is odd , ∴ P(n + 1) = 1

MATHS 

Integration : 

d 

If f(x) = F(x), then 

dx ∫ F(x) dx = f(x) + c, where c 

is an arbitrary constant called constant of integration. 

+ 1 

x n 

1. ∫ x dx 

n 

= (n ≠ –1) 

n + 1 

2. ∫ dx 

1 

= log x 

x 

3. ∫ e dx 

x 

= e x 

4. ∫ a dx 

x 

= 

x 

a 

log 

e 

a 

5. ∫ sin x dx = – cos x 

6. ∫ cos x dx = sin x 

2 

7. ∫ sec x dx = tan x 

8. ∫ cos x dx = – cot x 

ec 2 

9. ∫ sec x tan x dx = sec x 

10. ∫ cosec x cot x dx = – cosec x 

⎛ x π ⎞ 

11. ∫ sec x dx = log(sec x + tan x) = log tan ⎜ + ⎟ 

⎝ 2 4 ⎠ 

12. 

⎛ x ⎞ 

∫ cosec x dx = – log (cosec x + cot x) = log tan ⎜ ⎟ 

⎝ 2 ⎠ 

13. ∫ tan x dx = – log cos x 

14. ∫ cot x dx = log sin x 

15. ∫ − 2 2 

a 

dx 

x 

16. ∫ + 2 2 

a 

dx 

x 

17. ∫ − 2 2 

x 

x 

dx 

a 

= sin –1 

INTEGRATION 

x –1 x 

= – cos 

a a 

1 –1 x 

= tan = – 

a a a 

1 cot –1 

1 –1 x 

= sec = – 

a a a 

⎛ x ⎞ 

⎜ ⎟ 

⎝ a ⎠ 

1 cosec –1 

⎛ x ⎞ 

⎜ ⎟ 

⎝ a ⎠ 

18. ∫ − 2 2 


x 

1 

1 

a 

19. ∫ − 2 2 

a 

x 

20. ∫ − 2 2 

x 

dx 

a 

21. ∫ + 2 2 

x 

dx 

a 

22. ∫ − 2 2 

x 

1 

= log 

2a 

1 

dx = log 

2a 

= log 

⎧ 

⎨ 

⎩ 

= log 

⎧ 

⎨ 

⎩ 

1 

a dx = x 

2 

x − a 

, when x > a 

x + a 

+ 

a + x 

, when x < a 

a − x 

− a 

⎫ 

⎬ 

⎭ 

2 2 

x x = cos h –1 

+ 

+ a 

⎫ 

⎬ 

⎭ 

2 2 

x x = sin h –1 

2 

2 

a − x + 

2 

1 a 2 sin –1 

⎛ x ⎞ 

⎜ ⎟ 

⎝ a ⎠ 

⎛ x ⎞ 

⎜ ⎟ 

⎝ a ⎠ 

⎛ x ⎞ 

⎜ ⎟ 

⎝ a ⎠ 

23. ∫ − 2 2 1 

x a dx = x 

2 

2 2 

x − a 

1 2 

– a log 

⎧ 

⎨x 

+ 

2 ⎩ 

2 2 

x − a 

⎫ 

⎬ 

⎭ 

24. ∫ + 2 2 1 

x a dx = x 

2 

2 2 

x + a 

1 2 

+ a log 

⎧ 

⎨x 

+ 

2 ⎩ 

2 2 

x + a 

⎫ 

⎬ 

⎭ 

25. 

f ´( x) 

∫ dx = log f(x) 

f ( x) 

26. ∫ 

f ´( x) 

dx = 2 

f ( x) 

f (x) 

Integration by Decomposition into Sum : 

1. Trigonometrical transformations : For the 

integrations of the trigonometrical products such as 

sin 2 x, cos 2 x, sin 3 x, cos 3 2. 

x, sin ax cos bx, etc., they are 

expressed as the sum or difference of the sines and 

cosines of multiples of angles. 

Partial fractions : If the given function is in the 

form of fractions of two polynomials, then for its 

integration, decompose it into partial fractions (if 

possible). 

Integration of some special integrals : 

dx 

(i) ∫ 2 

ax + bx + c 

Mathematics Fundamentals 

This may be reduced to one of the forms of the above 

formulae (16), (18) or (19).

dx 

(ii) ∫ 2 

ax + bx + c 

This can be reduced to one of the forms of the above 

formulae (15), (20) or (21). 

(iii) ∫ + bx + c dx 

ax 2 

This can be reduced to one of the forms of the above 

formulae (22), (23) or (24). 

( px + q) 

dx ( px + q) 

dx 

(iv) ∫ , 

2 

ax + bx + c ∫ 2 

ax + bx + c 

For the evaluation of any of these integrals, put 

px + q = A {differentiation of (ax 2 + bx + c)} + B 

Find A and B by comparing the coefficients of like 

powers of x on the two sides. 

1. If k is a constant, then 

∫ dx 

k = kx and ∫ k f ( x) 

dx = k∫ f ( x) 

dx 

2. ∫ f ( x) 

± f ( x)} 

dx 

{ 1 2 = ∫ f 1 ( x) 

dx ± ∫ f 2 ( x) 

dx 

Some Proper Substitutions : 

1. ∫ f(ax + b) dx, ax + b = t 

2. ∫ f(axn + b)x n–1 dx, ax n + b = t 

3. ∫ f{φ(x)} φ´(x) dx, φ(x) = t 

4. 

f ´( x) 

∫ dx , f(x) = t 

f ( x) 

5. ∫ − 2 2 

x 

6. ∫ + 2 2 

x 

a dx, x = a sin θ or a cos θ 

a dx, x = a tan θ 

2 

2 

7. ∫ 

a − x 

2 2 

a + x 

dx, x 2 = a 2 cos 2θ 

8. ∫ a ± x dx, a ± x = t 2 

a − x 

9. ∫ dx, x = a cos 2θ 

a + x 

10. ∫ − 2 

2ax x dx, x = a(1 – cos θ) 

11. ∫ − 2 2 

a 

x dx, x = a sec θ 

Substitution for Some irrational Functions : 

dx 

1. ∫ , ax + b = t 

( px + q) 

ax + b 

2 

dx 

2. ∫ 2 

( px + q) 

ax + bx + c 

1 

, px + q = 

t 

dx 

3. ∫ , ax + b = t 

2 

( px + qx + r) 

ax + b 

2 

dx 

1 2 2 

4. ∫ , at first x = and then a + ct = z 

2 

2 

( px + r) 

ax + c 

t 

Some Important Integrals : 

dx 

⎛ x − α ⎞ 

1. To evaluate ∫ , 

( x − α)( 

x − β) 

∫ ⎜ ⎟ dx, 

⎝ β − x ⎠ 


∫ 

( x − α)( 

β − x) 

dx. Put x = α cos 2 θ + β sin 2 θ 

2. 

dx dx 

To evaluate ∫ , 

a + b cos x ∫ a + bsin 

x 

, 

dx 

∫ a + b cos x + c sin x 

⎛ x ⎞ 

⎛ 2 x ⎞ 

⎜2 

tan ⎟ 

⎜1− 

tan ⎟ 

Replace sin x = 

⎝ 2 ⎠ 

and cos x = 

⎝ 2 ⎠ 

⎛ 2 x ⎞ 

⎛ 2 x ⎞ 

⎜1+ 

tan ⎟ 

⎜1+ 

tan ⎟ 

⎝ 2 ⎠ 

⎝ 2 ⎠ 

x 

Then put tan = t. 

2 

3. 

p cos x + q sin x 

To evaluate ∫ dx 

a + b cos x + c sin x 

Put p cos x + q sin x = A(a + b cos x + c sin x) 

+ B. diff. of (a + b cos x + c sin x) + C 

A, B and C can be calculated by equating the 

coefficients of cos x, sin x and the constant terms. 

4. To evaluate 

dx 

∫ , 

2 

2 

a cos x + 2b 

sin x cos x + c sin x 

dx 

dx 

∫ , 

2 

a cos x + b ∫ 2 

a + bsin 

x 

In the above type of questions divide N r and D r by 

cos 2 x. The numerator will become sec 2 x and in the 

denominator we will have a quadratic equation in tan 

x (change sec 2 x into 1 + tan 2 x). 

Putting tan x = t the question will reduce to the form 

dt 

∫ 2 

at + bt + c 

5. Integration of rational function of the given form 

x + a 

(i) ∫ 4 2 

x + kx + a 

2 

2 

4 

x − a 

dx, (ii) ∫ 4 2 

x + kx + a 

2 

2 

4 

dx, where 

k is a constant, positive, negative or zero. 

These integrals can be obtained by dividing 

numerator and denominator by x 2 , then putting 

a 

x – 

x 

2 

a 

= t and x + 

x 

2 

= t respectively. 

Integration of Product of Two Functions : 

1. ∫ f1(x) f2(x) dx = f1(x) ∫ f2(x) dx – 

' [ ( f 1 ( x) 

f2 

( x) 

] dx 

∫ ∫ dx 

Proper choice of the first and second functions : 

Integration with the help of the above rule is called

integration by parts, In the above rule, there are two 

terms on R.H.S. and in both the terms integral of the 

second function is involve. Therefore in the product 

of two functions if one of the two functions is not 

directly integrable (e.g. log x, sin –1 x, cos –1 x, tan –1 x 

etc.) we take it as the first function and the remaining 

function is taken as the second function. If there is no 

other function, then unity is taken as the second 

function. If in the integral both the functions are 

easily integrable, then the first function is chosen in 

such a way that the derivative of the function is a 

simple functions and the function thus obtained under 

the integral sign is easily integrable than the original 

function. 

2. ∫ sin( bx + c) 

dx 

e ax 

= 

= 

a 

e 

2 

ax 

a 

+ b 

e ax 

2 

2 

+ b 

3. ∫ cos( bx + c) 

dx 

e ax 

= 

= 

a 

e 

2 

ax 

a 

+ b 

e ax 

2 

2 

+ b 

4. ∫ ekx {kf(x) + f '(x)} dx = e kx f(x) 

[a sin (bx + c) – b cos (bx + c)] 

2 

⎡ 

−1 

b ⎤ 

sin ⎢bx 

+ c − tan ⎥ 

⎣ 

a ⎦ 

[a cos (bx + c) + b sin(bx + c)] 

2 

⎡ 

−1 

b ⎤ 

cos ⎢bx 

+ c − tan ⎥ 

⎣ 

a ⎦ 

5. 

⎛ x ⎞ 

∫ log e x = x(logex – 1) = x loge ⎜ ⎟ 

⎝ e ⎠ 

Integration of Trigonometric Functions : 

1. To evaluate the integrals of the form 

I = ∫ sin m x cos n x dx, where m and n are rational 

numbers. 

(i) Substitute sin x = t, if n is odd; 

(ii) Substitute cos x = t, if m is odd; 

(iii) Substitute tan x = t, if m + n is a negative even 

integer; and 

1 1 

(iv) Substitute cot x = t, if (m + 1) + (n – 1) is an 

2 2 

integer. 

2. Integrals of the form ∫ R (sin x, cos x) dx, where R is 

a rational function of sin x and cos x, are transformed 

into integrals of a rational function by the substitution 

x 

tan = t, where –π < x < π. This is the so called 

2 

universal substitution. Sometimes it is more 

x 

convenient to make the substitution cot = t for 

2 

0 < x < 2π. 

The above substitution enables us to integrate any 

function of the form R (sin x, cos x). However, in 

practice, it sometimes leads to extremely complex 

rational functions. In some cases, the integral can be 

simplified by – 

(i) Substituting sin x = t, if the integral is of the form 

∫ R (sin x) cos x dx. 

(ii) Substituting cos x = t, if the integral is of the form 

∫ R (cos x) sin x dx. 

dt 

(iii) Substituting tan x = t, i.e. dx = , if the 

2 

1+ t 

integral is dependent only on tan x. 

Some Useful Integrals : 

dx 

1. (When a > b) ∫ a + b cos x 

2 

= tan 

2 2 

a − b 

–1 ⎡ − ⎤ 

⎢ tan ⎥ 

⎢⎣ 

+ 2⎥⎦ 

x a b 

+ c 

a b 

dx 

2. (When a < b) ∫ a + b cos x 

x 

b − a tan − a + b 

1 

= – log a 

2 2 

b − a 

x 

b − a tan + a + b 

a 

dx 

3. (when a = b) ∫ a + b cos x 

= 1 x 

tan + c 

a 2 

dx 

4. (When a > b) ∫ a + bsin 

x 


= 

2 

2 

a − b 

2 

tan –1 

dx 

5. (When a < b) ∫ a + bsin 

x 

= 

1 

2 

b − a 

2 

log 

⎧ ⎛ x ⎞ ⎫ 

⎪a 

tan⎜ 

⎟ + b ⎪ 

⎪ ⎝ 2 ⎠ ⎪ 

⎨ 

⎬ + c 

2 2 

⎪ a − b ⎪ 

⎪⎩ 

⎪⎭ 

⎛ x ⎞ 

a tan⎜ 

⎟ + b − 

⎝ 2 ⎠ 

⎛ x ⎞ 

a tan⎜ 

⎟ + b + 

⎝ 2 ⎠ 

b 

b 

2 

2 

− a 

− a 

dx 

6. (When a = b) ∫ a + bsin 

x 

= 1 

[tan x – sec x] + c 

a 

2 

2 

+ c

MATHS 

Functions with their Periods : 

Function Period 

sin (ax + b), cos (ax + b), sec (ax + b), 

cosec (ax + b) 

2π/a 

tan(ax + b), cot (ax + b) π/a 

|sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|, 

|cosec (ax + b)| 

π/a 

|tan (ax + b)|, |cot (ax + b)| π/2a 

Trigonometrical Equations with their General 

Solution: 

Trgonometrical equation General Solution 

sin θ = 0 θ = nπ 

cos θ = 0 θ = nπ + π/2 

tan θ = 0 θ = nπ 

sin θ = 1 θ = 2nπ + π/2 

cos θ = 1 θ = 2nπ 

sin θ = sin α θ = nπ + (–1) n α 

cos θ = cos α θ = 2nπ ± α 

tan θ = tan α θ = nπ + α 

sin 2 θ = sin 2 α θ = nπ ± α 

tan 2 θ = tan 2 α θ = nπ ± α 

cos 2 θ = cos 2 α θ = nπ ± α 

sin θ = sin α 

* 

cosθ 

= cosα 

sin θ = sin α 

* 

tan θ = tan α 

tan θ = tan α 

* 

cosθ 

= cosα 

TRIGONOMETRICAL 

EQUATION 

θ = 2nπ + α 

θ = 2nπ + α 

θ = 2nπ + α 

* If α be the least positive value of θ which satisfy 

two given trigonometrical equations, then the general 

value of θ will be 2nπ + α. 

Note : 

1. If while solving an equation we have to square it, 

then the roots found after squaring must be 

checked whether they satisfy the original equation 

or not. e.g. Let x = 3. Squaring, we get x 2 = 9, 

∴ x = 3 and – 3 but x = – 3 does not satisfy the 

original equation x = 3. 

2. Any value of x which makes both R.H.S. and 

L.H.S. equal will be a root but the value of x for 

which ∞ = ∞ will not be a solution as it is an 

indeterminate form. 

3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or 

y z 

y = z or both. But = ⇒ y = z only and not 

x x 

x = 0, as it will make ∞ = ∞. Similarly, if ay = az, 

then it will also imply y = z only as a ≠ 0 being a 

constant. 

Similarly, x + y = x + z ⇒ y = z and x – y = x – z 

⇒ y = z. Here we do not take x = 0 as in the 

above because x is an additive factor and not 

multiplicative factor. 

4. When cos θ = 0, then sin θ = 1 or –1. We have to 

verify which value of sin θ is to be chosen which 

⎛ 1 ⎞ 

satisfies the equation cos θ = 0 ⇒ θ = ⎜n 

+ ⎟ π 

⎝ 2 ⎠ 

If sin θ = 1, then obviously n = even. But if 

sin θ = –1, then n = odd. 

Similarly, when sin θ = 0, then θ = nπ and cos θ = 1 

or –1. 

If cos θ = 1, then n is even and if cos θ = –1, then 

n is odd. 

5. The equations a cos θ ± b sin θ = c are solved as 

follows : 

Put a = r cos α, b = r sin α so that r = 

and α = tan –1 b/a. 

Mathematics Fundamentals 


2 

a + b 

The given equation becomes 

r[cos θ cos α ± sin θ sin α] = c ; 

c c 

cos (θ ± α) = provided ≤ 1. 

r r 

Relation between the sides and the angle of a triangle: 

1. Sine formula : 

2

sin A 

a 

sin B 

= 

b 

= 

sin C 

c 

1 

= 

2R 

Where R is the radius of circumcircle of triangle 

ABC. 

2. Cosine formulae : 

2 2 2 

2 2 2 

b + c − a a + c − b 

cos A = 

, cos B = 

, 

2bc 

2ac 

2 2 2 

a + b − c 

cos C = 

2ab 

It should be remembered that, in a triangle ABC 

If ∠A = 60º, then b 2 + c 2 – a 2 = bc 

If ∠B = 60º, then a 2 + c 2 – b 2 = ac 

If ∠C = 60º, then a 2 + b 2 – c 2 = ab 

3. Projection formulae : 

a = b cos C + c cos B, b = c cos A + a cos C 

c = a cos B + b cos A 

Trigonometrical Ratios of the Half Angles of a Triangle: 

a + b + c 

If s = in triangle ABC, where a, b and c are 

2 

the lengths of sides of ∆ABC, then 

A s ( s − a) 

B s ( s − b) 

(a) cos = , cos = , 

2 bc 2 ac 

C 

cos = 

2 

A 

(b) sin = 

2 

C 

sin = 

2 

A 

(c) tan = 

2 

s ( s − c) 

ab 

( s − b)( 

s − c) 

bc 

( s − a)( 

s − b) 

ab 

( s − b)( 

s − c) 

, 

s( 

s − a) 

B 

, sin = 

2 

( s − a)( 

s − c) 

, 

ac 

B ( s − a)( 

s − c) 

C ( s − a)( 

s − b) 

tan = 

, tan 

2 s( 

s − b) 

2 s( 

s − c) 

Napier's Analogy : 

B − C b − c A C − A c − a B 

tan = cot , tan = cot 

2 b + c 2 2 c + a 2 

A − B a − b C 

tan = cot 

2 a + b 2 

Area of Triangle : 

1 1 1 

∆ = bc sin A= ca sin B = ab sin C 

2 2 2 

∆ = 

a sin B sin C b sinC 

sin A c sin Asin 

B 

= 

= 

2 sin( B + C) 

2 sin( C + A) 

2 sin( A + B) 

1 2 

1 2 

1 2 

2 

2 ∆ 

sin A = s( s − a)( 

s − b)( 

s − c) 

= 

bc 

bc 

2 ∆ 

2 ∆ 

Similarly sin B = & sin C = 

ca 

ab 

Some Important Results : 

A B 

1. tan tan = 

2 2 

s − c A B 

∴ cot cot = 

s 2 2 

A B c C c 

2. tan + tan = cot = (s – c) 

2 2 s 2 ∆ 

A B a − b 

3. tan – tan = (s – c) 

2 2 ∆ 

A B 

4. cot + cot = 

2 2 

A B 

tan + tan 

2 2 

A B 

tan tan 

2 2 


= 

s 

s − c 

c 

s − c 

cot C 

2 

5. Also note the following identities : 

Σ(p – q) = (p – q) + (q – r) + (r – p) = 0 

Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0 

Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0 

Solution of Triangles : 

1. Introduction : In a triangle, there are six 

elements viz. three sides and three angles. In 

plane geometry we have done that if three of the 

elements are given, at least one of which must be 

a side, then the other three elements can be 

uniquely determined. The procedure of 

determining unknown elements from the known 

elements is called solving a triangle. 

2. Solution of a right angled triangle : 

Case I. When two sides are given : Let the 

triangle be right angled at C. Then we can 

determine the remaining elements as given in the 

following table. 

Given Required 

(i) a, b 

(ii) a, c 

a 

tanA = , B = 90º – A, c = 

b 

a 

sin A 

a 

sinA = , b = c cos A, B = 90º – A 

c 

Case II. When a side and an acute angle are given – 

In this case, we can determine 

Given Required 

(i) a, A 

B = 90º – A, b = a cot A, c = 

a 

sin A 

(ii) c, A B = 90º – A, a = c sin A, b = c cos A

a 

PHYSICS 

Questions 1 to 9 are multiple choice questions. Each 

question has four choices (A), (B), (C) and (D), out of 

which ONLY ONE is correct. Mark your response in 

OMR sheet against the question number of that 

question. + 3 marks will be given for each correct 

answer and – 1 mark for each wrong answer. 

1. In the circuit shown, the cell is ideal, with 

e.m.f. = 15 V. Each resistance is of 3Ω. The potential 

difference across the capacitors in steady state – 

R=3Ω C=3µF 

R R 

+ – 

R R 

(A) 0 (B) 9 V 

15V 

(C) 12 V (D) 15 V 

2. In a young double slit apparatus the screen is rotated 

by 60º about an axis parallel to the slits. The slits 

separation is 3mm, slits to screen distance (i.e AB) is 4 

m & wavelength of light is 450 nm. The separation 

between the 3 rd dark fringe on the either side of B. 

60º 

A B 

screen 

Based on New Pattern 

IIT-JEE 2012 

XtraEdge Test Series # 10 

Time : 3 Hours 

Syllabus : 

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus 

Instructions : 

Section - I 

• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for 

correct answer and -1 mark for wrong answer. 

• Question 10 to 13 are Reason and Assertion type question with one correct answer. +3 marks will be awarded for 

correct answer and –1 mark for wrong answer. 

• Question 14 to 19 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong 

answer. 

• Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly 

matched answer (i.e. +1 marks for each correct row) and No Negative marks for wrong answer. 

(A) 6 mm (B) 8 mm (C) 4 3 mm (D) 2 3 mm 

3. A black body emits radiation at the rate P when its 

temperature is T. At this temperature the wavelength 

at which the radiation has maximum intensity is λ0. If 

at another temperature T' the power radiated is P' & 

λ 0 

wavelength at maximum intensity is 

2 

then – 

(A) P'T' = 32 PT (B) P'T' = 16 PT 

(C) P'T' = 8 PT (D) P'T' = 4 PT 

4. If two identical string are stretched such that there is 

fractional increase in their length, the fractional 

increase in length of first string is f and second string 

is 2f. Then the ratio of their fundamental frequency 

is. (Assume both obey the Hook Law i.e. tension ∝ 

elongation in string) - 

(A) 

1 

2 

1+ 

2f 

1+ 

f 

(B) 2 

1+ 

2f 

1+ 

f 


(C) 

1 

2 

1+ 

f 

1+ 

2f 

(D) 

1+ 

f 

1+ 

2f 

5. A infinite line charge of charge density λ lies along 

the x axis and let the surface of zero potential passes 

through (0, 5, 12) m. The potential at point 

(2, 3, –4) is – 

z 

V = 0 

(0,5,12) 

y

(A) 

(C) 

λ 

2πε0 

λ 

4πε 0 

13 

ln (B) 

5 

13 

ln (D) 

5 

2λ 

13 

ln 

πε 

0 3 

λ 13 

ln 

πε 

0 3 

6. A satellite is put in an orbit just above the earth 

atmosphere with a velocity 1 . 5 times the velocity 

for a circular orbit at that height. The initial velocity 

imparted is horizontal. What would be the maximum 

distance of satellite from earth when it is in the orbit- 

(A) 3R (B) 4R (C) 2 R (D) 5 R 

7. The density of a uniform rod with cross section area 

A is ρ, its specific heat capacity is C and the 

coefficient of its linear expansion is α. Calculate the 

amount of heat that should be added in order to 

increase the length of the rod by ∆l. 

2AρC∆l 

AρCα 

(A) 

(B) 

α 

∆l 

AρC∆l 

ρC∆l 

(C) 

(D) 

α 

2Aα 

8. For the system shown each spring has a stiffness of 

175 N/m. The mass of the pulleys may be neglected. 

The period of vertical oscillation of block – (mass of 

block is 28 kg) 

k k 

Block 

28 kg 

fricationless 

surface 

π π 

(A) 2 s (B) π 2 s (C) s (D) 

5 5 

9. A uniform elastic rod of cross section area A, natural 

length L and young modulus Y is placed on a smooth 

horizontal surface. Now two horizontal force (of 

magnitude F and 3F) directed along the length of rod 

and in opposite direction act at two of its ends as 

shown. After the rod has acquired steady state (i.e. no 

further extension take place), the extension of the rod 

will be – 

Elastic rod 

F 3F 

2F 

(A) L 

YA 

4F (B) L 

YA 

F 

(C) L 

YA 

3F 

(D) L 

2YA 

2 s 

This section contains 4 questions numbered 10 to 13, 

(Reason and Assertion type question). Each question 

contains Assertion (A) and Reason (R). Each question 

has 4 choices (A), (B), (C) and (D) out of which ONLY 

ONE is correct. Mark your response in OMR sheet 

against the question number of that question. +3 marks 

will be given for each correct answer and – 1 mark for 

each wrong answer. 

The following questions given below consist of an 

"Assertion" (A) and "Reason" (R) Type questions. Use 

the following Key to choose the appropriate answer. 

(A) If both (A) and (R) are true, and (R) is the correct 

explanation of (A). 

(B) If both (A) and (R) are true but (R) is not the 

correct explanation of (A). 

(C) If (A) is true but (R) is false. 

(D) If (A) is false but (R) is true. 

10. Assertion (A) : As the earth revolves around the sun 

it has an acceleration which is directed towards 

centre of the sun. 

Reason (R) : Angular momentum of the earth about 

the sun remains constant. 

11. Assertion (A) : In electric circuits , wires carrying 

currents in opposite direction are often twisted 

together. 

Reason (R) : The magnetic field in the surrounding 

space of a twisted wire system in not precisely zero. 

12. Assertion (A) : A metal ball is floating in mercury. 

Coefficient of volume expansion of metal is less than 

that of mercury. If temperature is increased, fraction 

of volume immersed of metal will increase. 

Reason (R) : Effect of heating on density of mercury 

will be more compared to that of metal. 

13. Assertion (A) : Work done by static friction is 

always zero. 

Reason (R) : Static friction acts when there is no 

relative motion between two bodies in contact. 

This section contains 2 paragraphs; each has 3 multiple 

choice questions. (Questions 14 to 19) Each question 






Passage # 1 (Ques. 14 to 16) 

A conducting rod PQ of mass M rotates without 

friction on a horizontal plane about Ο on circular 

rails of diameter 'l'. The centre O and the periphery 

are connected by resistance R. The system is located 

in a uniform magnetic field perpendicular to the 

plane of the loop. At t = 0, PQ starts rotating 

clockwise with angular velocity ω0. Neglect the 

resistance of the rails and rod, as well as self 

inductance. 


R 

P 

⊗ B 

O ω0 

14. Magnitude of current as a function of time 

2 

Bω0l −αt 

(A) e 

2R 

2 

Bω0l −2αt 

(B) e 

16R 

2 

Bω0l −αt 

(C) e 

8R 

2 2 

3B 

l 

Where α = 

8RM 

2 

Bω0l −2αt 

(D) e 

8R 

15. Total charge flow through resistance till rod PQ stop 

rotating . 

ω0M 

ω0M 

(A) (B) 

8B 

3B 

ω0M 

ω0M 

(C) (D) 

6B 

9B 

16. Heat generated in the circuit by t = ∞ 

M 

(A) 

24 

2 2 

l ω0 

M 

(B) 

8 

2 2 

l ω0 

M 

(C) 

3 

2 2 

l ω0 

M 

(D) 

32 

2 2 

l ω0 


A cylindrical spool of radius R is rigidly attached to a 

pulley of radius 2R. The mass of the combination is 

m and the radius of gyration about the centroidal axis 

G is R. A constant horizontal force F is applied at 

one end of the tape. Assume rolling motion between 

the pulley and the ground. I is the moment of inertia 

about centroidal axis. 

2R 

G R 

17. The acceleration of point P on the tape relative to the 

ground is - 

F 

(A) 

2m 

F 

(B) 

5m 

F 

(C) 

3m 

F 

(D) 

4m 

18. The length of the tape wound/unwound on the spool 

in time t is - 

Ft 

(A) 

10 m 

2 

Ft 

(B) 

5m 

2 

Ft 

(C) 

4m 

2 

Ft 

(D) 

2m 

2 

19. The linear acceleration of centre G is - 

2 

2FR 

(A) 

I 

2 

2FR 

(B) 

2 

I + mR 

2 

2FR 

(C) 

2 

I + 2mR 

2 

2FR 

(D) 

2 

I + 4mR 

P 

Q 

F 

This section contains 3 questions (Questions 20 to 22). 

Each question contains statements given in two columns 

which have to be matched. Statements (A, B, C, D) in 

Column I have to be matched with statements (P, Q, R, S) 

in Column II. The answers to these questions have to be 

appropriately bubbled as illustrated in the following 

example. If the correct matches are A-P, A-S, B-Q, B-R, 

C-P, C-Q and D-S, then the correctly bubbled 4 × 4 

matrix should be as follows : 

P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 

Mark your response in OMR sheet against the question 

number of that question in section-II. + 6 marks will be 

given for complete correct answer and No Negative 

marks for wrong answer. However, 1 mark will be 

given for a correctly marked answer in any row. 

20. A charged particle passes through a region that could 

have electric field only or magnetic field only or both 

electric and magnetic field or none of the fields. 

Match the possibilities 

Column-I 

(A) Kinetic energy of the particle remain constant. 

(B) Acceleration of the particle is zero 

(C) Kinetic energy of the particle changes and it also 

suffers deflection 

(D) Kinetic energy of the particle changes but it 

suffers no deflection 

Column-II 

(P) Under special condition this is possible when 

both electric and magnetic fields are present 

(Q) The region has electric field 

(R) The region has magnetic field only 

(S) The region contains no field 

21. Match the column : 

Column-I 

Phenomena on which machine work 

(A) Electromagnetic induction 

(B) Light of suitable frequency falling on a material 

result in emissions of electrons from the material 

(C) Change of orientation of a coil in a magnetic 

field results in e.m.f. across the coil 

(D) Mutual induction 

Column-II 

Machine or instrument 

(P) Photocell 

(Q) DC motor 

(R) AC generator 

(S) Transformer 


22. Match the column : 

Column-I 

(A) A photon stimulates the emission of another 

photon of 

(B) Photons of electromagnetic wave of different 

wavelengths may have 

(C) Two points on a wavefront must have 

(D) For constructive interference the waves must 

have 

Column-II 

(P) Same direction 

(Q) Same energy 

(R) Same phase 

(S) Same frequency 

CHEMISTRY 







1. The phenomenon of optical activity will be shown by - 

A 

A 

B 

B A 

(A) 

M 

B B 

A 

A 

(C) en M 

A 

en 

(B) 

(D) 

2. Titration curve of Na2CO3 and HCl is as given below. 

The indicators In1 and In2 respectively, must be – 

pH 

12 

10 

8 

6 

4 

2 

B 

B 

en 

M 

B 

M 

en 

10 20 30 40 50 60 

Volume of HCl (ml) 

B 

en 

In1 

In2 

(A) phenolphthalein and methyl orange 

(B) methyl orange and phenolphthalein 

(C) methyl orange and phenol red 

(D) phenolphthalein and phenol red. 

3. Metallic sulphates can be obtained by reacting the 

metals (above hydrogen in ECS), or its oxide, 

hydroxide or carbonate with dil.H2SO4. Gp IA 

metals also form hydrogen sulphates which can be 

isolated in solid. In general metal sulphates are 

soluble in water and crystallizes with water of 

crystallization. Sulphate are thermally more stable 

than nitrates. Select the stable hydrogen sulphate 

which can be obtained in solid state - 

(A) KHSO4 (B) CaHSO4 

(C) FeHSO4 (D) All of these 

4. Which of the following represent glyptal - 

O O 

(A) ––O–CH2–CH2–O–C C––– 

4 

O O 

(B) ––NH–(CH2)6–NH–C––(CH2)4– C––– 

4 

O O 

(C) 

O 

O 

4 


(D) 

O 

N 

H 

5. In the Cannizzaro's reaction given below : 

⎯ – 

OH 

2Ph – CHO ⎯→ ⎯ Ph–CH2OH + PhCOO –- 

the slowest step is - 

(A) the attack of OH – at the carbonyl group 

(B) the transfer of hydride to the carbonyl group 

(C) the abstraction of proton from the carboxylic acid 

(D) the deprotonation of Ph–CH2OH 

6. The relative rates of solvolysis in 80% EtOH of the 

following bromides is in the order – 

Br 

Br 

I 

II 

III 

(A) I > II > III (B) III > II > I 

(C) II > III > I (D) II > I > III 

4 

Br


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7. NaOH can be prepared by two methods each of the 

method has two steps having 100% yield. 

Method I : 2Na + 2H2O → 2NaOH + H2 

2H2 + O2 → 2H2O 

1 

Method II : 2Na + O2 → Na2O 

2 

Na2O + H2O → 2NaOH 

Which of the above method gives better yield of 

NaOH ? 

(A) Method I 

(B) Method II 

(C) Method I & Method II give equal yields. 

(D) Yield cannot be determined 

8. Which of the following value θ is correspond to the 

maximum dipole moment of the triatomic molecule 

XY2 

Y 

θ 

X 

(A) θ = 90º (B) θ = 120º (C) θ = 150º (D) 180º 

9. Photons having energy equivalent to binding energy 

of 2 nd state of Li + ion are used at metal surface of 

work function 10.6 eV. If the ejected electrons are 

further accelerated through the potential difference of 

5 V then the minimum value of de-Broglie 

wavelength associated with the electron is – 

(A) 2.45 Å (B) 9.15 Å (C) 5 Å (D) 11 Å 


















10. Assertion (A) : 

N 

O O N 

O O 

is a macrocyclic ligand 

O 

O 

Reason (R) : The ligand in which donor atoms are N 

and O is known a cryptands. 

Y 

11. Assertion (A) : For adsorption ∆G, ∆S and ∆H all have 

negative values. 

Reason (R) : Adsorption is spontaneous process 

accompained by increase in entropy. 

12. Assertion (A) : For the concentration cell, 

Zn(s) | Zn 2+ (aq)(C1) | | Zn 2+ (aq) (C2)/Zn, for 

spontaneous cell reaction C1 < C2. 

RT C2 

Reason (R) : For concentration cell,Ecell = ln 

nF C 

For spontaneous reaction, Ecell +ve ⇒ C2 > C1 

13. Assertion (A) : Aryl halides undergo nucleophilic 

substitution with ease. 

Reason (R) : Carbon-halogen bond in aryl halides 

has partial double bond character. 









FeCl3 on reaction with K4 [Fe(CN) 6 ] in aqueous 

solution gives blue colour, according the 

equation 

4FeCl3 + 3K4[Fe(CN)6] → Fe4[Fe(CN)6]3 + 12 KCl 

At 300 K two aqueous solution of K4 [Fe(CN) 6 ] & 

FeCl3 with equal concentrations 0.1 M. These two 

solutions are separated by a semipermeable 

membrane AB as shown in figure. 

A 

0.1M 0.1M 

K4[Fe(CN)6] 

B 

14. What is correct – 

(A) side ‘x’ is hypotonic 

(B) side ‘y’ is hypotonic 

(C) both are isotonic 

(D) None of these 

15. What is true about the solutions – 

(A) blue colour forms in side ‘x’ 

(B) blue colour forms in side ‘y’ 

(C) blue colour forms on both sides 

(D) No blue colour formation 

16. By applying external pressure osmosis can be 

stopped it should be applied to – 

(A) side ‘x’ (B) side ‘y’ 

(C) equal on both the sides 

(D) can not be stopped by applying external pressure 


side'x' 

FeCl3 

side'y' 

1


An activating group activates all positions of the 

benzene ring; even the positions meta to it are more 

reactive than any single position in benzene itself. It 

directs ortho and para simply because it activates the 

ortho and para positions much more than it does the 

meta. 

A deactivating group deactivates all positions in the 

ring, even the positions meta to it. It directs meta 

simply because it deactivates the ortho and para 

positions even more than it does the meta. Thus, both 

ortho and para orientation and meta orientation arise 

in the same way : The effect of any group whether 

activating or deactivating is strongest at the ortho and 

para positions. 

But certain groups (– NH2 and – OH, and their 

derivatives) act as powerful activators towards 

electrophilic aromatic substitution, even though they 

contain electronegative atoms and can be shown in 

other ways to have electron-withdrawing inductive 

effects. 

Halogens are unusual in their effect on electrophilic 

aromatic substitution; they are deactivating yet ortho, 

para-directing. Deactivation is characteristic of 

electron withdrawal, whereas ortho-para orientation 

is characteristic of electron release. 

17. Which will undergo Friedel-Crafts alkylation 

reaction? 

(1) 

(3) 

CH3 

NO2 

COOH 

(2) 

(4) 

CH2CH3 

OH 

(A) 1, 2 and 4 (B) 1 and 3 

(C) 2 and 4 (D) 1 and 2 

18. Which of the following is the strongest acid ? 

OH 

(A) OH 

(C) 

OH 

NO2 

NO2 

(B) 

(D) 

Cl 

OH 

NO2 

19. Reactivity in halogen substituted benzene rings is 

controlled by : 

(A) resonance 

(B) inductive effect 

(C) inductive effect dominates resonance effect 

(D) resonance effect dominates inductive effect 










P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 






20. One mole of an ideal gas is subjected to the following 

process : 

Column-I Column-II 

(A) ∆W = 0 (P) If the gas undergoes 

free expansion 

(B) ∆E = 75Cv (Q) If the gas is cooled 

reversibly at constant 

pressure from 373 K 

to 298 K 

(C) ∆H = –75 Cp (R) If the gas is heated 

from 298 K to 373 K 

reversively at constant 

pressure 

(D) ∆W = –75 R (S) If the gas is heated 

from 298 K to 373 K at 

constant volume 

21. Match the Column : 


(Gases X and Y (Tatio of times 

taken of diffusion) taken) 

(A) X = 100 ml of H2 at 1 bar, 25º C (P) 1 : 1225 

Y = 200 ml of O2 at 2 bar, 25ºC 

(B) X = 100 ml of O2 at 1 bar, 25ºC (Q) 1 : 0.7 



(C) X = 100 ml of SO2 at 1 bar, 25ºC (R) 1 : 1.36 


(D) X = HCl gas to travel 100 cm (S) 1 : 8 

length in a tube 

Y = NH3 gas to travel 200 cm 

length using the same tube 

(P, V, T) = same in both cases) 

22. Match the Column : 


(A) Soluble in dil HCl (P) SrS2O3 

(B) Yields SO2 with dil (Q) SrSO3 

HCl on bonding 

(C) Soluble in aqueous (R) SrSO4 

solution 

(D) Insoluble in aqueous (S) CdS 

solution 

MATHEMATICS 







1. The area enclosed by | y | – | x | = 1 and x 2 + y 2 =1 is 

(A) 2 units 2 (B) zero units 2 

(C) infinite units 2 (D) none of these 

2. Least value of the expression 9sec 2 θ + 4cosec 2 θ, is- 

(A) 6 (B) 1 (C) 36 (D) 25 

x 

3. Let f (x) = ∫ ( 2( 

x −1)( 

x − 2) 

+ 3( 

x −1) 

( x − 2) 

) dx , 

1 

then- 

(A) f has exactly 4 critical points 

(B) f has maximum at x = 2 

(C) x = 7/5 is minima & x = 1 is maxima 

(D) none of these 

4. The locus of the middle points of chords of a 

parabola which subtend a right angle at the vertex of 

the parabola is- 

(A) Circle (B) Parabola 

(C) Ellipse (D) Straight line 

5. The probability that a particular day in the month of 

july is a rainy day is 3/4. Two person whose 

credibility are 4/5 and 2/3 respectively claim that 15 th 

july was a rainy day. The probability that it was real 

a rainy day. 

(A) 3/4 (B) 24/25 (C) 8/9 (D) none 

3 

2 

2 

6. 

− ⎛ 2 −[ 

] ⎞ 

Domain of f (x) = sin ⎜ ⎟ 

⎝ [ ] ⎠ 

1 x x 

, where [.] denotes 

x 

the greatest integer function, is 

(A) (–∞, 1) – {0} 

⎡ 4 ⎞ 

(B) ⎢− 

, 0⎟ 

∪ {0} 

⎣ 3 ⎠ 

(C) (–∞, 0) ∪ I + (D) (–∞, ∞) – [0, 1) 

7. The number of different words of three letters which 

can be formed from the word "PROPOSAL", if a 

vowel is always in the middle are- 

(A) 53 (B) 52 

(C) 63 (D) 32 

8. Let a1, a2, a3, ...... be terms of an A.P. If 

a1 

+ a2 

+ .... + a p 

a1 

+ a2 

+ ..... + aq 

2 

p a6 

= , p ≠ q, then 

2 

q 

a21 

equals- 

(A) 41/11 (B) 7/2 

(C) 2/7 (D) 11/41 

9. The curve y = ax 3 + bx 2 + cx is inclined by 45º to 

x-axis at origin and it touches x-axis at (1,0). Then- 

(A) a = –2, b = 1, c = 1 (B) a = 1, b = 1, c = –2 

(C) a = 1, b = –2, c = 1 (D) a = –1, b = 2, c = 1 


















10. Assertion (A) : Let z be a complex number satisfying 

|z – 3| ≤ |z – 1|, |z – 3| ≤ |z – 5|, |z – i| ≤ |z + i| 

and |z – i| ≤ |z – 5i|. Then the area of region in which 

z lies is 12 sq. units. 

1 

Reason (R) : Area of trapezium = (sum of parallel 

2 

sides) (Distance between parallel sides) 

11. Let f (x) = | 1 – x | and g(x) = sin –1 (f | x |) 

Assertion (A) : Number of values of x, where g(x) is 

non differentiable is 3. 

Reason (R) : Domain of g(x) is [–1, 1] 


12. Assertion (A) : If eccentricities of two ellipse are 

same then their areas are also same. 

2 2 

x y 

Reason (R) : Area of the ellipse + = 1 

2 2 

a b 

(a < b, a > 0, b > 0) is π ab square units. 

13. Consider a circle S : (x – 2) 2 + (y – 3) 2 = 13 and a line 

L : y = x – 12. 

Assertion (A) : Chord of contact of pair of tangents 

drawn from every point on L = 0 to S = 0 passes 

through P(3, 2) 

Reason (R) : Pole of polar L = 0 with respect to 

S = 0 is P(3, 2) 









If f (xy) = f (x) . f (y) and f is differentiable at x = 1 

such that f '(1) = 1 also f (1) ≠ 0, then 

14. f (x) is - 

(A) continuous for all x ∈ R 

(B) discontinuous at x = –1, 0, 1 

(C) differentiable for all x ≠ 0 


15. f '(7) equals- 

(A) 7 (B) 14 (C) 1 (D) None 

16. Area bounded by curve f(x), x axis and ordinate 

x = 4, is- 

(A) 64/3 (B) 8 (C) 16 (D) None 


1 

There exists a G.P. with first term and common 

A 

1 

ratio A (A > 1). If we add in the sum of first n 

2 

terms of the sequence, it equals to the sum of the 

coefficients of even power of x in the expansion of 

(1 + x) n . If we interchange the first term & common 

ratio of given G.P., the sum of new infinitely 

decreasing G.P. is equal to B, where A, B and n are 

related by the relation 

B−2 

∫ 

A−2 

n 364 

( 1+ 

x) 

dx = 

3 

17. 

( 1+ 

x) 

− n − B 

The value of lim 

isx→A 

x − A 

(A) 3 (B) 6 (C) e (D) 8 

A 

18. Area bounded by f(x) = x A and g(x) = x B is- 

A + B 

(A) 

n 

2A 

(C) 

A + 2B 

+ n 


(B) 

(D) 

B − A 

n 

B 

n + A + B 

19. Number of real roots of the equation 

(x B – nx A ) 1/A = 6 are 

(A) 2 (B) 4 (C) 1 (D) 0 










P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 








(A) The reflection of the point (P) 5 

(t – 1, 2t + 2) in a line is 

(2t + 1, t) then the line has 

slope equal to 

(B) If θ be the angle between (Q) 6 

two tangents which are drawn 

to the circle x 2 + y 2 – 6 3 x – 6y + 27 = 0 

from the origin, then 2 3 tanθ 

equals to 

(C) The shortest distance between (R) 2 7 

parabolas y 2 = 4x and y 2 = 2x – 6 

is d then d 2 = 

(D) Distance between foci of the (S) 1 

curve represented by the equation 

x = 1 + 4cosθ, y = 2 + 3sinθ is

21. Column-I Column-II 

(A) If y = 2[x] + 9 = 3[x + 2], where (P) – 1 

[.] denotes greatest integer function, 

1 

then [x + y] is equal to 

6 

(B) If 

⎛ 1 1 ⎞ 

lim ⎜sin 

+ cos ⎟ 

⎠ 

x→∞⎝ x x 

k is equal to 

(C) If three successive terms of a G.P. (R) 2 

with common ratio r, (r > 1) forms 

the sides of a triangle then [r] + [–r] 

is equal to (where [.] denotes 

greatest integer function) 

(D) Let f(x) = (x 2 – 3x + 2)(x 2 +3x+2) (S) 3 

and α, β, γ are the roots of 

f '(x) = 0, then [α]+[β] + [γ] is 

equal to (where [.] denotes greatest 

integer function) 

x 

= e k/2 then (Q) 0 

22. Column-I Column-II 

(A) The order and degree of the 

differential equation 

2 

dy d y 

3 − 4 − 7x 

= 0 are 

dx 2 

dx 

(P) 13 

a and b then a + b is 

(B) If kˆ r 

a = î + 2jˆ 

+ 3 , kˆ r 

b = 2î 

− jˆ + 

and k 

(Q) 102 

ˆ 

r 

r r r 

c = 3î 

+ 2jˆ 

+ and a × ( b× 

c) 

r r r 

is equal to xa 

+ yb 

+ zc 

, then 

x + y + z is equal to 

(C) The number of 4 digit numbers 

that can be made with the digits 

1,2,3,4,3,2 

(R) 5 

dx 

(D) If ∫ = 

2 2 

( x + 1)( 

x + 4) 

(S) 7 

a −1 

a −1⎛ 

x ⎞ 

tan x − tan ⎜ ⎟ + k 

b c ⎝ d ⎠ 

where k is constant of integration, 

then 2a + b + c + d is 

(where a & b and a & c are 

co-prime numbers) 

Chemistry Facts 

• At 0 degress Celsius and 1 atmospheric pressure, 

one mole of any gas occupies approximately 

22.4 liters. 

• Atomic weight is the mass of an atom relative to 

the mass of an atom of carbon-12 which has an 

atomic weight of exactly 12.00000 amu. 

• If the atom were the size of a pixel (or the size 

of a period), humans would be a thousand miles 

tall. 

• It would require about 100 million 

(100,000,000) atoms to form a straight line one 

centimeter long. 

• The weight (or mass) of a proton is 

1,836.1526675 times heavier than the weight (or 

mass) of an electron. 

• The electron was first discovered before the 

proton and neutron, in 1897 from English 

physicist John Joseph Thomson. 

• The neutron was discovered after the proton in 

1932 from British physicist James Chadwick, 

which proved an important discovery in the 

development of nuclear reactors. 

• Carbon dioxide was discovered by Scottish 

chemist Joseph Black. 

• When silver nitrate is exposed to light, it results 

in a blackening effect. (Discovered by Scheele, 

which became an important discovery for the 

development of photography). 


PHYSICS 







1. Ice point on a particular scale measure 22º. The scale 

agrees with Fahrenheit scale at 72ºF. The 

temperature in Fahrenheit scale when new scale 

reads 27º is - 

(A) 40ºF (B) 38ºF (C) 36ºF (D) 34ºF 

2. A hot body is kept in a chamber maintained at fixed 

lower temperature. Time taken by body in loosing 

half the maximum heat it can lose is 

3. 

5 min. Time taken by body in loosing the maximum 

heat it can lose is - 

(A) 10 min (B) 12 min (C) 20 min (D) Infinite 

A point source of power 4 W is placed at the centre 

of spherical shell. The shell is kept in a vacuum 

chamber maintained at 27ºC. If shell attains a 

constant temperature 37ºC, emissivity of surface of 

shell is nearly - 

1 

(A) 

2 

1 

(B) 

3 

2 

(C) 

3 

4 

(D) 

5 

4. In a resonance–column experiment, a long tube, open 

at top, is clamped vertically. Water level inside tube 

can be moved up or down. First resonance is 

occuring when water level is at depth 20 cm below 

open end. Let second resonance occurs when water 

level is at a distance ‘x’ below opening, then - 

Based on New Pattern 

IIT-JEE 2013 

XtraEdge Test Series # 10 

Time : 3 Hours 

Syllabus : 

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus 

Instructions : 

• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for 

correct answer and -1 mark for wrong answer. 

• Question 10 to 13 are Reason and Assertion type question with one correct answer. +3 marks will be awarded for 

correct answer and –1 mark for wrong answer. 

• Question 14 to 19 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong 

answer. 

• Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly 

matched answer (i.e. +1 marks for each correct row) and No Negative marks for wrong answer. 

(A) x = 40 cm (B) x = 60 cm 

(C) x < 60 cm (D) x > 60 cm 

5. Figure shows a toy-whistle. It a disk made of plastic 

having two conical grooves at diametrically opposite 

point. When it is rotated about its centre with 

sufficiently high speed air intercepted by groove 

produce whistling sound. An observer is at a distance 

10 m from the centre of toy. Radius of toy is 15 cm. 

Frequency of sound emitted by toy when it is rotating 

1800 

with ω = rpm is 10 kHz. If velocity of sound 

π 

in air 300 m/s, beat frequency heard by observer is – 

(A) 10 Hz (B) 15 Hz (C) 20 Hz (D) None 

6. An ice cube is fixed at the bottom of container 

containing water. Water in the container will be 

cooled majorly by - 

(A) Convection (B) Conduction 

(C) Radiation (D) Convection and conduction 

7. In a region of space a constant force F newton acts 

on a particle of mass m, which is released from rest 

at point A. When the particle reaches B its – 

m 

A B 


→ 

F 

O

(A) potential energy (PE) increases but kinetic energy 

(KE) decreases. 

(B) PE decreases but KE increases. 

(C) PE remains constant but KE increases. 

(D) PE decreases but KE remains constant 

8. A L shaped rod whose one rod is horizontal and other 

is vertical is rotating about a vertical axis as shown 

with angular speed ω. The sleeve shown in figure has 

mass m and friction coefficient between rod and 

sleeve is µ. The minimum angular speed ω for which 

sleeve cannot sleep on rod is – 

(A) 

(C) 

ω = 

ω = 

µ l 

ω 

l 

m 

g (B) 

sleeve 

ω = 

µ g 

l 

l 


µ g 

9. Four particles of equal mass M move along a circle of 

radius R under the action of their mutual gravitational 

attraction. The speed of each particle is - 

GM 

(A) 

R 

(B) 

⎡ 

⎢2 

⎣ 

GM ⎤ 

2 

R 

⎥ 

⎦ 

⎡GM 

⎤ 

(C) ⎢ ( 2 2 + 1) 

⎥ 

⎣ R ⎦ 

(D) 

⎡GM 

⎛ ⎞⎤ 

⎢ ⎜ 

2 2 + 1 

⎟⎥ 

⎢ 

⎜ ⎟ 

⎣ 

R ⎝ 4 ⎠⎥⎦ 


















10. Assertion (A) : A string of length 28 cm, fixed at one 

end vibrates in its 3 rd overtone. Wave motion will not 

be disturbed by holding the string at a distance 8 cm 

from fixed end. 

Reason (R) : Position of nodes in the above string 

will at distance 4 cm, 8 cm, …. from fixed end. 

11. Assertion (A) : Two tube of same length but 

different diameter vibrating in same harmonic. 

Frequency of tube having smaller diameter will be 

more. 

Reason (R) : End-correction is less in tube of smaller 

diameter. 

12. Assertion (A) : When an observer moves towards 

stationary source frequency heard by observer is 

more than that frequency emitted by source. 

Reason (R) : Wavelength of sound wave received by 

observer becomes smaller. 

13. Assertion (A) : At height h from ground and at depth 

h below ground, where h is approximately equal to 

0.62 R, the value of g acceleration due to gravity is 

same. 

Reason (R) : Value of g decreases both sides, in 

going up and down. 









A vertical cylindrical tube has its lower end closed. 

There is small opening in the bottom of the tube by 

which level of water in tube can be adjusted. Length 

of tube is 50 cm. Velocity of sound in air and water 

is 320 m/s and 1440 m/s respectively. Water is filled 

in tube upto height 30 cm. 

14. Minimum frequency with which tube can resonate is- 

(A) 300 Hz (B) 320 Hz 

(C) 400 Hz (D) 450 Hz 

15. Minimum frequency so that standing waves can be 

formed both in air and liquid column - 

(A) 400 Hz (B) 800 Hz 

(C) 640 Hz (D) 900 Hz 

16. Air column in tube is resonating with a tuning fork at 

its lowest possible frequency. Water level is lowered 

slowly. Minimum distance by which water level has 

to be lowered so that intensity of sound become 

maximum is - 

(A) 10 cm (B) 20 cm 

(C) 40 cm (D) 50 cm 


A wire of length 1 m is clamped horizontally between 

two rigid support. 

17. Wire is plucked at a distance 30 cm from one end. 

Lowest harmonic in which wire can resonate is - 

(A) 3rd (B) 4th (C) 5th (D) 10th 


18. A light ring which can slip frictionlessly over wire is 

slipped over wire. When a tuning fork start 

resonating with wire, ring is found to be at rest at a 

distance 15 cm from one end. If velocity of wave in 

wire is 10 m/s, minimum frequency of tuning fork is- 

(A) 50 Hz (B) 60 Hz 

(C) 70 Hz (D) None of these 

19. Now two light rings are slipped on wire which can 

move frictionlessly. When wire is plucked, the two 

rings are found to be at rest separated by 12 cm. 

Maximum wavelength of wave in wire is - 

(A) 4 cm (B) 8 cm 

(C) 12 cm (D) 24 cm 










P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 






20. A small ring of mass m passes through a smooth wire 

bent in form of horizontal circle. The ring is 

connected to a spring whose other end is fixed at A 

on the wire as shown. The natural length of spring is 

mg 

R and spring constant is where m is mass of the 

R 

ring and R is also the radius of circle. Initially the 

ring is released from rest from position B and it 

moves towards C as in the figure. (N = Normal 

reaction between wire and ring, v = speed of ring) 

C 

wire 

A 

60º 


(A) N = mg (P) at position B 

(B) N = zero (Q) at position C 

(C) N = 2 mg (R) some where between 

position B and position C 

(D) v = gR (S) never 

B 

21. A body of mass ‘m’ is acted upon by net force F r . 

r and v be their initial position and velocity. All 

quantity are in S.I. unit . Then match the following. 


2 

(A) F = −4x 

î , r = 0 , (P) Motion : S.H.M. 

v = 2î, 

m = 2 

(B) j ˆ 

F = ( 2 − 4y) 

, r = 0 (Q) Motion : Non-periodic 

v = 0 , m = 1 

3 

(C) F = − 2( 

x + x ) î , (R) Path : Straight line 


– 2 

r = 10 î v = 0 , m = 1 

(D) F = − 4( 

xî 

+ yjˆ 

) 

j 

(S) Time period : π 

ˆ r = 2 , v = 6î, 

m = 1 

22. The cubical container filled with water is given 

acceleration → 

a = a0 î + a0 j ˆ + a0 k ˆ , then : (neglect 

the effect of gravity) 

E F 

C 

H 

D 

G 

B 

A z 


(A) Pressure at point A is (P) less than pressure 

at point G 

(B) Pressure at point D is (Q) less than pressure 

at point F 

(C) Pressure at point E is (R) Greater than 

pressure at point C 

(D) Pressure at point H is (S) Greater than pressure 

at point B 

CHEMISTRY 







1. Which of the substance has maximum number of 

hydrogen atoms per gm of the substance ? 

At. wts. : Cu – 63.5 ; S – 32 ; O –16 ; H –1 

(A) CH4 (B) CuSO4 . 5H2O 

(C) H2O2 (D) H2O 

y 

x

2. In the reaction 3 Cu + 8 HNO3 → 3 Cu(NO3)2 

+ 2NO + 4 H2O 

what is the equivalent weight of HNO3? if molecular 

weight of HNO3 is M - 

M 3 

(A) M (B) (C) M 

3 

4 

4 

(D) M 

3 

3. The order of magnitude of ionic radii of ions Na + , 

Mg 2+ , Al 3+ and Si 4+ is : 

(A) Na + > Si 4+ > Al 3+ > Mg 2+ 

(B) Mg 2+ > Na + > Al 3+ > Si 4+ 

(C) Na + > Mg 2+ > Al 3+ > Si 4+ 

(D) Si 4+ > Al 3+ > Mg 2+ > Na + 

4. The correct order of tendency of polymerization is - 

(A) SiO4 4– < PO4 3– < SO4 2– < ClO4 – 

(B) PO4 3– < SiO4 4– < SO4 2– < ClO4 – 

(C) ClO4 – < SO4 2– < SiO4 4– < PO4 3– 

(D) SiO4 4– > PO4 3– > SO4 2– > ClO4 – 

5. 

K P for the gaseous reaction – 

K 

C 

(a) 2 A + 3 B 2C 

(b) 2 A 4B 

(c) A + B + 2C 4D 

would be respectively - 

(A) (RT) –3 , (RT) 2 , (RT)º 

(B) (RT) –3 , (RT) –2 , (RT) –1 

(C) (RT) –3 , (RT) 2 , (RT) 

(D) None of the above 

6. No. of heteroatoms (other than C) present in the 

following heterocyclic compound is – 

O 

N–H 

O 

(A) 3 (B) 2 (C) 1 (D) 0 

7. & are – 

(A) Tautomers (B) Functional 

(C) Position (D) All the above 

8. Which does not exists in solid state - 

(A) LiHCO3 (B) CaCO3 (C) NaHCO3 (D) Na2CO3 9. Decomposition of H2O2 is retarded by - 

(A) Acetanilide (B) MnO 2 

(C) Zinc (D) Finely divided metals 


















10. Assertion (A) : F atom has a less negative electron 

gain enthalpy than Cl atom. 

Reason (R) : Additional electron are repelled more 

effectively by 3p electron in Cl atom than by 2p 

electron in F atom. 

11. Assertion (A) : When a non-volatile solute is added 

to water ice equilibrium, some ice melts. 

Reason (R) : Ice melts to dilute the solution in order 

to increase the vapour pressure in accordance to 

Le-Chatelier's principle. 

12. Assertion (A) : 

cyclohexancarbonitrile. 

H 

CN 

is called 

Reason (R) : It is an aromatic compound. 

13. Assertion (A) : The following compounds are 

optically inactive. 

CH3 

H Cl 

H Cl 

CH3 

CH3 

H Cl 

CH2 

H Cl 


CH3 

H 

CH3 

CH2 

CH3 

Cl 

CH2 

H Cl 

CH3 

H Cl 

CH2 

CH2 

CH2 

H Cl 

CH3 

Reason (R) : the meso compounds do not have any 

chiral C atom so have optical rotation is zero. 







each wrong answer.


PCl5 in solid state exists as PCl4 + and PCl6 – . Also in 

some solvents it undergoes dissociation as 

2PCl5 PCl4 + + PCl6 – 

14. The geometry and hybridisation of PCl5 is - 

(A) Trigonal bipyramid, sp 3 d 

(B) Tetrahedral, sp 3 

(C) Octahedral, sp 3 d 2 

(D) none of these 

15. The geometry and hybridisation of PCl4 + is - 

(A) Tetrahedral, sp 3 

(B) Octahedral, sp 3 d 2 

(C) Trigonal pyramid, sp 3 d 

(D) See-saw, sp 3 d 

16. The geometry and hybridisation of PCl6 – is - 

(A) Octahedral, sp 3 d 2 

(B) Tetrahedral, sp 3 d 2 

(C) Square planar bipyramid, sp 3 d 

(D) See-saw, sp 3 d 


The IE 1 and IE II in kJ/mol of a few elements are 

given in the following table 

Element IEI in kJ/Mol IEII in kJ//mol 

P 2372 5251 

Q 520 7300 

R 900 1760 

S 1680 3380 

17. Which of the above element is likely to be alkali 

metal ? 

(A) P (B) Q (C) R (D) S 

18. Which of the above element is likely to be alkaline 

earth metal ? 

(A) P (B) Q (C) R (D) S 

19. Which of the above element is likely to be noble gas? 

(A) P (B) Q (C) R (D) S 










P Q R S 

A P Q R S 

B P Q R S 

C P Q R S 

D P Q R S 







(A) Primary standard base (P) Na2C2O4 

Molecular wt 

(B) Equiv. wt = 

(Q) Na2B4O7 

2 

(C) Primary standard reducing (R) Na2CO3 

agent 

(D) Capable of reducing 

hardness of water due to 

the presence of Ca 

(S) As2O3 

2+ 

21. Column-I Column-II 

(A) Tc/Pc (P) Z 

(B) Tc/Vc (Q) a/Rb 

(C) TB (R) 8b/R 

(D) Ti/TB (S) 8a/81 Rb 2 



(A) 

(B) 

(C) 

P A 

V 

T 

B 

(P) 

P A B (Q) Isotherm 

P 

A 

B 

T 

Temperature 

is increasing 

(R) Isochoric

(D) 

P 

A 

T 

B 

(S) Pressure is increasing 

MATHEMATICS 







1. In a ∆ABC, if r = r2 + r3 –r1 and ∠A > π/3 then range 

s 

of is equal toa 

⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 

(A) ⎜ , 2⎟ 

(B) ⎜ , ∞⎟ 

(C) ⎜ , 3⎟ 

(D) (3, ∞) 

⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 

2. Vector perpendicular to î – j ˆ – k ˆ and in the plane of 

î + j ˆ + kˆ and – î + j ˆ + kˆ is 

(A) î + kˆ (B) 2 î + j ˆ + kˆ (C) 3 î +2 j ˆ + k ˆ (D) 4 î – 2 j ˆ –2 k ˆ 

3. If a, b ∈ R, a ≠ 0 and roots of ax 2 – bx + 1 = 0 

imaginary, then a + b + 1 is 

(A) Zero (B) Positive 

(C) Negative (D) None of these 

4. Number of distinct normals that can be drawn to the 

curve x 2 = 4y from point (1, 2), is 

(A) 0 (B) 1 (C) 2 (D) 3 

5. ABC is a triangle whose medians AD and BE are 

perpendicular to each other. If AD = p and BE = q 

then area of ∆ABC is- 

2 

(A) pq 

3 

3 

(B) pq 

2 

4 

(C) pq 

3 

3 

(D) pq 

4 

6. If (21.4) a = (0.00214) b =100, then the value of 

1 1 

– is : 

a b 

(A) rational but not integral 

(B) prime 

(C) irrational 

(D) composite 

7. For 3 ≤ r ≤ n 

n 

Cr + 3 n Cr–1 + 3 n Cr–2 + n Cr–3 is- 

(A) n+3 Cr (B) 2 n+2 Cr+2 (C) 3 n+1 Cr+1 (D) 3 n Cr 

8. Point of intersection of straight lines represented by 

6x 2 + xy – 40y 2 – 35x – 83y + 11 = 0 is- 

(A) (3, 1) (B) (3, –1) (C) (–3, 1) (D) (–3, –1) 

9. Number of points outside the hyperbola 

3x 2 – y 2 = 48 from where two perpendicular tangents 

can be drawn to the hyperbola is/are - 

(A) 1 (B) 2 (C) infinite (D) None 


















10. Assertion (A) : The length of the shortest intercept 

made by the family of lines (1 + λ) x + (λ – 1) y 

+ 2 (1– λ) = 0 on the parabola x 2 = 4(y – 1) is 5. 

Reason (R) : Latus rectum is the shortest focal chord 

of the parabola. 

11. Assertion (A) : If S1 and S2 are non-concentric 

circles then their radical axis must exist. 

Reason (R) : S1,S2, S3 are three circles such that no two 

are concentric then their radical centre is defined. 

12. Assertion (A) : If two straight lines intersect the xaxis 

at A and B and y-axis at C and D such that 

OA.OB = OC.OD, O being origin then points A, B, 

C, D are concyclic. 

Reason (R) : If a secant through a point P intersects 

a circle at Q and R then PQ.PR is independent of the 

direction of the secant. 

13. Assertion (A) : The equation 

(log x) 2 – log x 3 + 2 = 0 has only one solution. 

Reason (R) : log x 2 = 2 log x if x > 0 










It is given that A = (tan –1 x) 3 + (cot –1 x) 3 , where 

x > 0 & B = (cos –1 t) 2 + (sin –1 t) 2 , where 

⎡ 1 ⎤ 

t∈ ⎢0 

, ⎥ & sin 

⎣ 2 ⎦ 

–1 x + cos –1 π 

x = for –1 ≤ x ≤ 1 & 

2 

π 

∀ x∈R 

tan –1 x + cot –1 x = 2 

14. The interval in which A lie is 

⎡ 3 3 

π π ⎞ ⎡ 3 3 

(A) ⎢ , ⎟ 

π π ⎞ ⎛ 3 3 ⎞ 

⎢ 

⎟ 

(B) ⎢ , ⎟ 

⎣ 8 2 

⎟ 

(C) ⎜ 

π π 

⎟ 

⎠ ⎢⎣ 

32 8 ⎜ 

, 

⎟ 

(D) None 

⎠ ⎝ 10 5 ⎠ 

15. Maximum value of B is 

2 

π 

(A) 

8 

2 

π 

(B) 

16 

2 

π 

(C) 

4 

(D) None 

16. If least value of A is λ & max. value of B is µ then 

cot –1 ⎛ ⎛ λ − µπ ⎞⎞ 

⎜ 

⎟ 

⎜ 

cot ⎜ ⎟ 

⎟ 

= 

⎝ ⎝ µ ⎠⎠ 

π − π 7π −7π 

(A) (B) (C) (D) 

8 8 

8 

8 


Consider lines 

x − 2 y − 3 z 4 

L1 : = = 

1 1 − k 

− , L2 : 2 

x − 1 y − 4 z − 5 

= = 

2 1 

17. A vector perpendicular to L1 &L2 and of length 3 2 

is- (When k = 1) 

(A) 3 î –3 j ˆ (B) 2(3 î +2 j ˆ – kˆ ) 

(C) –3 î +4 j ˆ + 2 kˆ (D) 3 î + 4 kˆ 18. Value of 'k' so that lines L1 and L2 are coplanar, is - 

(A) –1 (B) –1/2 (C) –2 (D) 2 

19. Equation of plane containing these lines is 

(A) x – y – 2 = 0 (B) 2x – y + 2 = 0 

(C) x – y + 7 = 0 (D) None of these 










P Q R S 

A 

B 

C 

D 

P Q R S 

P Q R S 

P Q R S 

P Q R S 






20. Match the following 

Consider a plane P = 0 on whom foot of 

perpendicular from point (1, 1, 1) is (2, 3, 4). 


(A) sum of intercepts of P = 0 

on coordinate axis is 

(P) 52/7 

(B) perpendicular distance of (Q) 110/3 

(0, 0, 0) from plane is λ then λ 2 is 

(C) A line through (0, 0, 0) and (R) 120 

perpendicular to plane is 

x y z 

= = then a + b+ c may be 

a b c 

(D) Radius of circle obtained by (S) 200/7 

plane 'P' and sphere x 2 + y 2 + z 2 

= 36 is 'r' then r 2 21. 

is equal to 

Match the Column : 


(A) Distance of 3x + 4y – 5 = 0 

from (1, 1) is 

(P)1 

(B) Area of ∆ formed by x + y = 

with co-ordinate axis is 

2 (Q) 4/3 

(C) Circumcentre of ∆ formed by 

3x + 4y – 7 = 0 and axis is 

(h, k) then h + k is 

(R) 2/5 

(D) Two sides of ∆ are x + y = 1 (S) 49/24 

and 2x + y + 4 = 0. If circumcentre 

is (2, 1) then slope of third side is 



(A) Find the number of 6 digit (P) 1 

natural numbers, where each 

digit appears at least twice 

(B) In how many ways can five (Q) 677 

different books be tied up in 

three bundles 

(C) How many non-empty (R) 11754 

collections are possible by 

using 5P's and 6 Q's 

(D) How many students do you (S) 25 

need in a school to guarantee 

that there are atleast 2 students, 

who have the same 1st two 

initials in their 1st names 



PHYSICS 

1. The spectrum of hydrogen atom has many lines 

although hydrogen atom contain only one electron 

why ? 

2. What happens to the frequency when light passes 

from one medium to another. 

3. What are coherent sources of light 

4. If the intensity of the incident radiation on a metal is 

doubled, what happens to the kinetic energy of the 

emitted photoelectrons. 

5. Prove mathematically that the average power over a 

complete cycle of alternating current through an 

ideal inductor is zero. 

6. Figure shows a branch of a circuit. Find out the value 

of potential difference between A and B. 

A 

1A 

2Ω + – 

2V 

1Ω 

B 

7. Write the following radiations in an ascending order 

in respect of their frequencies : 

X–rays, microwaves, ultraviolet rays, radiowaves 

MOCK TEST PAPER-3 

CBSE BOARD PATTERN 

CLASS # XII 

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS 

Solutions will be published in same issue 

General Instructions : Physics & Chemistry 

• Time given for each subject paper is 3 hrs and Max. marks 70 for each. 

• All questions are compulsory. 

• Marks for each question are indicated against it. 

• Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. 

• Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. 

• Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. 

• Question numbers 28 to 30 are long-answer questions and carry 5 marks each. 

• Use of calculators is not permitted. 

General Instructions : Mathematics 

• Time given to solve this subject paper is 3 hrs and Max. marks 100. 

• All questions are compulsory. 

• The question paper consists of 29 questions divided into three sections A, B and C. 

Section A comprises of 10 questions of one mark each. 

Section B comprises of 12 questions of four marks each. 

Section C comprises of 7 questions of six marks each. 

• All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 

• There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. 

• Use of calculators is not permitted. 

8. In a closed surface incoming flux and outgoing flux 

is 5 × 10 5 and 4 × 10 5 V-m respectively then find out 

the charge enclosed by surface. 

9. A freshly prepared radioactive substance of half life 

2 hours emits radiation of intensity which is 64 times 

the permissible safe level. Find the minimum time 

after which it would possible to work with this 

source safely. 

10. Why does the conductivity of a semiconductor change 

with the rise in temperature. 

11. In optical fiber refractive index of cladding is less 

than core. Why ? 

12. Where should an object be placed from a convex lens 

to from an image of the same size ? Can it happen in 

the case of a concave lens ? 

13. A transmitting antenna at the top of tower has a 

height of 36 m and the height of the receiving 

antenna is 49 m. What is the maximum distance 

between them for satisfactory communication in the 

LOS mode ? 

(Radius of earth = 6400 km) 


14. Derive the expression for the capacitance of a 

parallel plate capacitor, having two identical plates 

each of area A and separated by a distance d, when 

the space between the plates is filled by a dielectric 

medium. 

15. In a meter bridge shown in circuit diagram, the 

balance point is found to be at 40 cm from the 

end A when resistor Y is of value 20 Ω. 

Calculate the shifting in of balancing point on the 

same wire if X and Y are interchanged. 

X Y 

A 

D 

B 

G 

E 

16. Two parallel wires of infinite length have linear 

charge densities λ1 and λ2 coulomb/metre. Derive an 

expression for force per unit length acting between 

them. 

17. A 10 µF capacitor is charged by a 30 V d.c. supply 

and then connected across an uncharged 50 µF 

capacitor. Calculate (i) the final potential difference 

across the combination and (ii) the initial and final 

energies. How will you account for the difference in 

energy ? 

18. Amplitude of electric field in an electromagnetic wave 

is 3×10 6 V/m. Find amplitude of magnetic field. 

19. Explain it 

(i) work function 

(ii) Threshold frequency 

(iii) Threshold wave length 

(iv) Stopping potential 

20. Show that the energy of the first excited state of He + 

atom is equal to the energy of the ground state of 

hydrogen atom 

21. In a young's double slit experiment, the slits are 

separated by 0.56 mm and the screen is placed 2.8 m 

away. The distance between the central bright fringe 

and the fifth bright fringe is measured to be 1.5 cm. 

Determine the wavelength of light used in the 

experiment. 

22. A communication system having operating wavelength 

λ in m can use only x% of its source frequency as its 

channel bandwidth. The system is to be used for 

transmitting TV signals requiring bandwidth of F Hz, 

How many channels can this system transmit 

simultaneously 

C 

23. The given figure shows a network of resistances R1, 

R2, R3 and R4. Using Kirchhoff's laws, establish the 

balance condition for the network. 

B 

A G C 


R1 

R2 

R3 R4 

D 

+ – 

24. Distinguish between resistance, reactance and 

impedance for an a.c. circuit. Draw graph showing 

variation for reactance of (i) a capacitor (ii) an 

inductor with frequency of the circuit. 

25. A 0.5 m long metal rod PQ completes the circuit as 

shown. The area of the circuit is perpendicular to the 

magnetic field of flux density 0.15 T. If the 

resistance of the total circuit is 3 Ω. Calculate the 

force needed to move the rod in the direction as 

indicated with speed 2 ms –1 . 

× × × × × × 

Q 

× × × × 

× × × × × × × × × × 

× × × × × × × × × v × 

× × × × × × × × × × 

× × × × × × P × × × × 

26. Explain the principle and working of cyclotron with 

the help of a labelled diagram. 

27. Using Gauss' theorem, deduce an expression for the 

electric field intensity at any point due to a thin, 

infinitely long wire of charge/length λ C/m. 

28. What is diffraction ? write down its difference with 

interference ? If single slit diffraction is obtained by 

using light of wavelength 5000Å and slit width 0.5 

mm. Then calculate angular width of central 

maxima. 

OR 

What is dispersion ? How the angular dispersion is 

defined ? Which colours is deviated maximum by a 

prism ? Calculate dispersive power of crown glass if 

refractive indices for red, yellow and violet colours 

are 1.5140, 1.5170 and 1.5318 respectively. 

• 29. What is spectrum ? write down types of spectrum 

? 

• OR 

• If electrons trans it from first excited state of 

H-atom to ground state and emitted radiations 

incidents over a metallic surface of threshold 

wavelength 4000 Å, then what is the value of

maximum kinetic energy of electrons emitted from 

this surface ? 

30. With the help of a neat and labelled diagram, explain 

the underlying principle and working of a moving 

coil galvanometer. What is the function of: 

(i) uniform radial field 

(ii) soft iron core in such a device ? 

OR 

Derive a mathematical expression for the force per 

unit length experienced by each of the two long 

current carrying conductors placed parallel to each 

other in air. Hence define one ampere of current. 

Explain why two parallel straight conductors carrying 

current in the opposite direction kept near each other in 

air repel ? 

CHEMISTRY 

1. Complete the following reactions : 

CH3–CH2–CH=CH2 + HCl → …………. 

2. Arrange the following compounds in an increasing 

order of their acid strengths : 

(i) Br2 – CH2 – CH2 – COOH 

(ii) Cl – CH – CH2 – COOH 

| 

Cl 

(iii) F – CH2 – CH2 – COOH 

(iv) CH2 – CH2 – COOH 

| 

I 

3. How will you distinguish between propanal & 

propanone. 

4. What are amino acid. Write zwitterionic structure of 

alanine. 

5. What happen when an electric field is applied to a 

colloidal solution ? 

6. Explain the Frenkel defect with suitable example. 

7. Why is the elevation in b.p. of water different in the 

following solutions ? 

(i) 0.1 molar NaCl solution. 

(ii) 0.1 molar sugar solution. 

8. Write a IUPAC name of following : 

CH3 – CH – C – C – OCH3 

Br O O 

9. Complete the following reaction sequense : 

(i) C6H5N2Cl + Cu2Cl2 ⎯⎯→ ∆ 

…………… 

(ii) HC ≡ CH 

Hg−SO4 ⎯ ⎯ 

H2SO4 

⎯ → 

.................. 

10. For the following conversion reactions write the 

chemical equations : 

(i) Ethyl isocyanide to ethylamine 

(ii) Aniline to N-Phenylethanamide 

11. Write the structural formulae of A, B, C and D in the 

following sequence of reaction : 

A + CH3MgBr H2O CH3CH2 – CH – CH3 

OH 


D 

Alc. KOH 

C 

Br2 

B 

–H2O 

H2SO4 

12. Give the structural formulae and names of the 

products of the following reactions : 

(i) Chloroform is heated with aniline in presence of 

alc. KOH. 

(ii) Phenol is treated with an excess of aqueous 

bromine. 

13. For a I st order chemical reaction rate constant is 

2.303 sec –1 . Determine the time required for 90% 

completion of a chemical reaction 

14. The vapour pressure of pure liquid A and B are 450 

mm Hg and 700 mm Hg at 350 K respectively. Find 

the mole fractions of the components if total vapour 

pressure is 600 mm Hg. 

15. Calculate the mass of a nonvolatile solute (molecular 

mass 40) which should be dissolved in 114 g octane 

to reduce its vaour pressure to 80%. 

16. Describe the following with an example for each : 

(i) Aldol condensation 

(ii) Trans-esterification 

17. Write reactions starting conditions for the following 

conversions : 

(i) Benzene to Acetophenone 

(ii) Ethanal to Propanone 

18. (a) Compare the magnetic moments of the following 

Cr +3 , V +3 , Fe +3 

(b) Which of the following is coloured 

CuSO4(aq), [Sc(H2O)6] +3 , ZnCl2 

19. How are the following conversions carried out ? 

(Write reactions only) 

(i) 1-bromopropane to 2-bromopropane 

(ii) Propanone to iodoform 

(iii) Phenol to salicylic acid 

20. Describe in brief the structure of DNA. 

21. Write the structure of polymers : 

(a) PVC (b) PMMA (C) PTFE 

22. What are antacids. Explain with example 

23. What is adsorption ? How does adsorption of a gas 

on a solid surface vary with

(a) temperature and (b) Pressure ? 

Illustrate with the help of appropriate graphs. 

24. Explain the following terms with suitable examples : 

(i) Schottky defect, (ii) Interstitials 

(iii) F-centres. 

25. A first order reaction is 20% complete in 10 minutes. 

Calculate (i) k of the reaction and (ii) time taken for 

the reaction to go upto 75% completion. 

26. (a) Name the lanthanoide element which forms 

tetrapositive ions in aqueous solution. 

(b) Why do Zr and Hf exhibit similar properties ? 

(c) Colour of a solution of K2Cr2O7 depends on pH 

of the solution. Why ? 

27. (a) Write preparation of K2Cr2O7. Give reactions. 

(b) Complete reactions : 

K2Cr2O7 + H2SO4 + KI ⎯→ 

KMnO4 + H2SO4 + H2S ⎯→ 

28. (a) Describe the preparation of acetic acid from 

acetylene. 

(b) How can the following be obtained from acetic 

acid 

(i) Acetone 

(ii)Acetaldehyde 

(c) In what way can acetic acid be distinguished 

from acetone ? 

(d) Why do carboxylic acids not give the 

characteristic reactions of a carbonyl group ? 

• OR 

• (a) How would you account for the following : 

• (i) Aldehydes are more reactive than ketones 

towards nucleophiles. 

(ii) The boiling points of aldehydes and ketones 

are lower than of the corresponding acids 

• (iii) The aldehydes and ketones undergo a 

number of addition reactions. 

• (b) Give chemical tests to distinguish between : 

(i) Acetaldehyde and benzaldehyde 

• 

• 

(ii) Propanone and propanol 

• 29. (a) Calculate the EMF of the cell Mg (s)/Mg 2+ 

(0.2 M) || Ag + (1 × 10 –3 M) / Ag ; Mg / Mg 

2 Θ 

E + = – 

Θ 

2.37 V, E + 

Ag / Ag = + 0.80 V. What will be the 

effect on EMF if concentration of Mg 2+ ion is 

decreased to 0.1 M ? 

• (b) (i) Arrange the following metals in the order in 

which they displace each other from the 

solution of their salts. 

(ii) Given the standard electrode potentials ; 

K + / K = –2.93 V, Ag + / Ag = 0.80 V, 

Hg 2+ / Hg = + 0.79 V, Mg 2+ / Mg = – 2.37 V, 

Cr 3+ / Cr = – 0.74 V 

• Arrange these metals in their increasing order 

of reducing power. 

30. Arrange the following in order of the property 

mentioned and give reason. 

(i) F2, Cl2, Br2, I2 (Bond energy) 

(ii) HCl, HF, HI, HBr (acidic strength) 

(iii) PH3, NH3, SbH3, AsH3 (basic strength) 

(iv) H2O, H2S, H2Se, H2Te (boiling point) 

(v) HOCl, HClO2, HClO3, HClO4 (acidic strength) 

MATHEMATICS 

Section A 

1. From the differential equation representing the 

family of curves y = A cos (x + B), where A and B 

are parameter. 

2. Evaluate : ∫ 1 + sin x / 2 dx . 

3. If → 

a = iˆ 

– 2 ˆj 

+ 3kˆ 

and → 

b = kˆ î – 3 , find 

| → 

b × 2 → 

a |. 

4. If → 

a = kˆ î + 2 jˆ – 3 ; → 

b = 3 iˆ 

– ˆj 

+ 2kˆ 

, show that 

( → 

a + → 

b ) is perpendicular to ( → 

a – → 

b ). 

5. Show that the four points whose position vectors are 

6 iˆ 

– 7 ˆj 

, 16 iˆ 

– 29 ˆj 

– 4kˆ 

, 3 ˆj 

– 6kˆ 

and 

2 iˆ 

+ 5 ˆj 

+ 10kˆ 

are coplanar. 

6. Write the following function in the simplest form 

tan –1 x 

2 2 

a – x 

. 

7. Find gof : If f (x) = |x| and g(x) = |5x – 2|. 

8. If matrix A has order 3 × 4, matrix B has order 

4 × 2 and matrix C has order 2 × 3, Find order of 

(A.B.C). 

9. If A = 

find x. 

⎡ 0 

⎢ 

⎣ 2x 

– 3 

x + 2⎤ 

0 

⎥ 

⎦ 

10. Find value of x such that 

is skew symmetric, 


2 

5 

4 

1 

= 

2x 

6 

4 

. 

x

Section B 

11. The odds against a certain event are 5 to 2 and the 

odds in favour of another independent event are 6 to 

5. Find the chance that at least one of the events will 

happen. 

OR 

In two successive throws of a pair of dice, determine 

the probability of getting a total of 8, each time. 

12. Solve (1 + x 2 dy 2 

) +2xy – 4x = 0. 

dx 

13. Evaluate : ∫ ⎥ ⎡ + x ⎤ 

e ⎢ dx 

⎢⎣ 

+ x ⎦ 

x 1 

. 

2 

( 2 ) 

OR 

Evaluate∫ 2x 

2 4 

1− 

x − x 

dx 

dx 

14. Evaluate : ∫ . 

(sin x – 2cos 

x)( 

2sin 

x + cos x) 

15. The position vectors of two points A and B are 

3 iˆ 

+ ˆj 

+ 2kˆ 

and iˆ 

– 2 ˆj 

– 4kˆ 

respectively. Find the 

vector equation of the plane passing through B and 

perpendicular to the vector AB. 

16. If → 

a = iˆ 

– 3 ˆj 

+ kˆ 

, → 

b = iˆ 

– ˆj 

+ kˆ 

and 

→ 

c = 2 iˆ 

– ˆj 

+ kˆ 

, verify that → 

a × ( → 

b × → 

c ) = 

( → 

a . → 

c ) → 

b – ( → 

a . → 

b ) → 

c . 

17. Find the equation of tangent to curve y = 3x – 2 . 

Which is parallel to line 4x – 2y + 5 = 0. 

18. If y = 

dy 

= 

dx 

x 

..... ∝ 

x 

x , then show that 

2 

y 

. 

x( 

2 – y log x) 

Differentiate tan –1 

⎡ 

⎢ 

⎢ 

⎣ 

OR 

1+ 

x 

1+ 

x 

19. If x y = e x–y , then prove that 

dy 

= 

dx 

log x 

2 

( 1+ 

log x) 

2 

2 

− 

+ 

1− 

x 

1− 

x 

2 

2 

⎤ 

⎥ 

⎥ 

⎦ 

w.r.t,cos –1 x 2 

⎛ π ⎞ 

tan⎜ – x⎟ 

4 

20. If f (x) = 

⎝ ⎠ 

, x ≠ π/4, find the value which 

cot 2x 

can be assigned to f (x), at x = π/4 so that f (x) 

becomes continuous every where in [0, π/2]. 

21. Find the value of parameter α for which 

f (x) = 1 + αx, α ≠ 0 is the inverse of itself. 

22. Use properties of determinants proof that 

x + y x x 

5x 

+ 4y 

10x 

+ 8y 

Show that 


4x 

8x 

1 

1 

1 

x 

y 

z 

x 

x + 4 

and hence factorize. 

yz 

zx 

xy 

OR 

= 

= x 3 . 

1 

1 

1 

Section C 

23. One bag contains five white and four black ball. 

Another bag contains seven white and nine black 

balls. A ball is transferred from the first bag to the 

second and then a ball is drawn from the second. 

Find the probability that the ball drawn is white. 

24. Find the area of the region bounded by x 2 + y 2 = 1 and 

(x – 1) 2 + y 2 = 1. 

π 

∫ 

0 

cos x 

e 

25. Evaluate : dx . 

cos x – cos x 

e + e 

π/ 

4 

Or 

x + π / 4 

Evaluate : ∫ dx . 

2 – cos 2x 

– π/ 

4 

26. Find the shortest distance between the following 

lines : → 

r = (1 – t) î + (t – 2) jˆ + (3 – 2t) k ˆ and 

→ 

r = (s + 1) î + (2s – 1) jˆ – (2s + 1) k ˆ . 

27. A farmer has a supply of chemical fertilizer of type A 

which contains 10% nitrogen and 6% phosphoric acid 

and of type B which contains 5% of nitrogen and 10% 

of phosphoric acid. After soil testing it is found that at 

least 7 kg of nitrogen and the same quantity of 

phosphoric acid is required for a good crop. the 

fertilizer of type A costs Rs. 5.00 per kg and the type 

B costs Rs. 8.00 per kg. Using linear programming 

find how many kgs of each type of the fertilizer 

x 

y 

z 

x 

y 

z 

2 

2 

2

should be bought to meet the requirement and the cost 

be minimum. Solve the problem graphically. 

OR 

A furniture dealer deals only in two items-tables and 

chairs. He has Rs.10,000 to invest and a space to 

store at most 60 pieces. A table costs him Rs. 500 

and a chair Rs.200. He can sell a table at a profit of 

Rs.50 and a chair at a profit of Rs.15. Assume that 

he can sell all items that he buys. Using linear 

programming formulate the problem for maximum 

profit and solve it graphically. 

28. A wire of length 36 m is to be cut into two pieces. 

One of the pieces is to be made into a square and 

other into a circle. What should be lengths of two 

pieces, so that combined area of the squared and 

circle is minimum. 

29. Solve the system of equation by using matrix 

method. 

2 3 10 4 6 5 

+ + = 4 ; 

– + = 1 ; 

x y z 

x y z 

6 9 20 

+ – = 2 

x y z 

Brain Teaser 

Pears - There are a few trees in a garden. On one of 

them, a pear tree, there are pears (quite logical). But 

after a strong wind blew, there were neither pears on 

the tree nor on the ground. How come? 

Pears – Solution At first, there were 2 pears on the 

tree. After the wind blew, one pear fell on the ground. 

So there where no pears on the tree and there were no 

pears on the ground.Another possible solutions: The 

wind blew so hard that the pears fell of the tree and 

blew along the ground into the water or hovering in 

the air in a tornado. 

1. Emeralds have been produced synthetically in 

labs since 1848 and can be virtually 

indistinguishable from the genuine article. 

2. In the last 200 years the use of metals has 

increased as scientists have discovered new ones: 

until the 17th Century only 12 metals were 

known - there are now 86. 

3. The only person to have an element named after 

him while still alive was Glenn Seaborg, the 

most prolific of all the element hunters. 

4. Traffic lights with red and green gas lights were 

first introduced in London in 1868. 

Unfortunately, they exploded and killed a 

policeman. The first successful system was 

installed in Cleveland, Ohio in 1914. 

5. In 1998, design student Damini Kumar at South 

Bank University patented a teapot with a special 

grooved spout, which she claims virtually rules 

out dribbling. 

6. Even though most items in the home today are 

technologically up to date, most of us are still 

using the standard light bulb designed in 1928! 

7. A chest x-ray is comprised of 90,000 to 130,000 

electron volts. 

8. The strength of early lasers was measured in 

Gillettes, the number of blue razor blades a given 

beam could puncture. 

9. The first commercial radio station in the United 

States, KDKA Pittsburgh, began broadcasting in 

November 1920. 

10. A British rocket attack on US soldiers is 

celebrated in the lyrics of the US National 

Anthem. 

11. Until the late 1800s, it was forbidden for women 

in the United States to obtain a patent, so if a 

woman had invented something she would file 

for a patent under her husband or father's name. 

For this reason, the number of early female 

inventors remains a mystery. 

XtraEdge for IIT-JEE 73 

12. Milt Garland, a 102 year old engineer, invented a 

technology that forms ice on the exterior of a 

casing instead of inside it, which is used to create 

indoor ice rinks. 

FEBRUARY 2012

PHYSICS 

1. Refractive index is maximum for violet and 

minimum for red colour 

2. Path difference between two waves should be odd 

multiple of λ/2 or phase difference between two 

waves should be odd multiple of π. 

3. 

P 

−34 

h 6. 

6 × 10 

= = 


−10 

–27 kg-m/s 

λ 

5000 × 10 

4. R = R0A 1/3 ⇒ R ∝ A 1/3 

∴ 

R 1 

= 

R 2 

⎛ A 

⎜ 

⎝ A 

3R 

∴ R2 = 

2 

1 

2 

⎞ 

⎟ 

⎠ 

1 

3 

⇒ 

R 

= 

R 2 

1 

⎛ 8 ⎞ 

⎜ ⎟⎠ 

⎝ 27 

/ 3 

2 

= 

3 

5. It should be very high i.e. in KHz to GHz 

6. 

= 

µ 

c 

v ; v2 < v3 < v1 

7. Resistivity or specific resistance of a material is 

defined as the resistance offered by a wire made by 

that material having unit area of cross-section and 

unit length. 

W 

8. V = = 

q 

MOCK TEST-2 (SOLUTION) 


6× 

10 

3. 

2× 

10 

– 5 

– 7 

MOCK TEST– 2 PUBLISHED IN JANUARY ISSUE 

= 50 volt 

9. Objective lens must have greater focal length than 

eye-piece because magnifying power of lens 

⎛ f ⎞ 

⎜ 0 ⎟ 

⎜ 

M P = 

⎟ 

increases in this case. 

⎝ fe 

⎠ 

10. Conditions of interference 

(i) waves must travel in the same medium in the 

same direction. 

(ii) Their frequencies and wavelengths must be 

same 

(iii) Their plane of polarisation must be same 

(iv) Their must be maintained a constant phase 

difference between the waves. 

11. Ionisation energy:- Minimum energy required to 

remove electron from its ground state. 

Excitation energy:- Energy required to transit an 

electron to one of its excited state from ground state 

KE = 

2r 

; U = – 

∴ U = −2KE 


KZe 2 

12. Emax. = 12V; Emin. = 4V 

Ma = 

E 

E 

max 

max 

− E 

+ E 

min 

min 

KZe 2 

r 

12 − 4 

= = 

12 + 4 

1 

2 

1 

∴ ma in percentage = × 100 = 50% 

2 

13. Modulating signal being of low frequency cannot 

travel to large distance so it is modulated with high 

frequency carrier. Frequency modulation is virtually 

free from noise where as amplitude modulation 

suffers from noise pollution. 

14. Semiconductor atoms are tetravalent. The form 

covalent bonds by sharing their electrons with the 

neighboring atoms. When donor impurity of valency 

S is doped, the donor atoms takes the place of a 

semiconductor atom in the crystal lattice. Donor 

atoms forms covalent bond by sharing its four 

electrons whereas the fifth electron of donor atom is 

so loosely attached to the atom such that it behaves 

as a free electron at room temperature. 

15. (i) width of depletion layer of zener diode 

becomes very small due to heavy doping of 

p and n regions. 

(ii) The field across depletion layer becomes very 

⎛ V ⎞ 

high ⎜Q 

E = ⎟ 

⎝ d ⎠ 

16. Let at any instant charge on the capacitor is q, then 

q 

p.d. V = 

C 

Then work done to put additional charge dq is 

⎛ q ⎞ 

dW = (dq) V = ⎜ ⎟ dq 

⎝ C ⎠ 

∴ Total work done to give charge Q is 

Q 

⎛ q ⎞ Q 

W = ∫⎜ ⎟dq = 

⎝ C ⎠ 2C 

2 

1 2 

= CV (Q Q = CV) 

2 

0

This work done is stored inside the capacitor as P.E. 

1 2 

∴ U = W = CV 

2 

17. (i) A galvanometer can be converted into ammeter 

by connecting a low resistance called shunt parallel 

to the galvanometer coil. 

(ii) A galvanometer can be converted into voltmeter 

by connecting a large resistance in series with the 

galvanometer coil. 

18. (i) Electromagnetic waves are produced by 

accelerated charge. 

(ii) The velocity of electromagnetic wave in free 

space is equal to the velocity of light in free space. 

(iii) It is transverse in nature. 

19. α = 2º 

β = ? 

fo = 20 cm 

fe = 4 cm 

β 

MP = and MP = 

α 

⎛ 20 ⎞ 

∴ β = ⎜ ⎟ × 2 = 10 

⎝ 4 ⎠ 

20. 


f 

f 

o 

e 

∴ 

β 

α 

f 

= 

f 

h 

λ = 

2mqV 

for proton 

h 

λ = 

…(i) 

2meV 

For α-particle 

h 

λ = 

…(ii) 

2× 

4m× 

2eV' 

By (i) and (ii) 

2meV = 2 × 4m × 2eV' 

V '= 

V / 8 

BE/A 

e 

d 

c 

b 

a 

Fe 56 

o 

e 

⎛ f 

⇒ β = ⎜ 

⎝ f 

A 

o 

e 

⎞ 

⎟ 

α 

⎠ 

a → 2He 4 

b → 4Be 8 

c → 6C 12 

d → 8O 16 

e → 10Ne 20 

22. 2 is NAND and 1 is OR gate 

A B Y' Y 

0 1 1 0 

23. R = (AB × 10 c ± 5)% 

Yellow A = 4 

Violet B = 7 

Brown C = 1 

∴ R = (47 × 10 ± 5)% 

= (470 ± 5)% 

24. (i) To make strong electromagnet 

(ii) To make high speed computers 

25. tanθ = 


B 

B 

V 

H 

∴ BV = BH tanθ = 0.4 × 10 –4 tan60º = 0.69 × 10 –4 T and 

BH = Be cos θ 

⇒ Be = 

−4 

BH 

0. 

4× 

10 

= 

cosθ 

cos60º 

= 0.8 × 10 –4 T 

26. Biot–Savart′s law 

This law is applicable to determine magnetic field 

due to small current element. Magnetic field due to 

small element → 

d l at point A is given as 

i 

→ 

dl → 

dl 

→ 

r 

→ → 

→ µ 0i 

( dl× 

r 

= 

3 

dB 

4π 

µ 0 idl 

sin θ 

or dB = 

4π 

2 

r 

where µ0 ⇒ Permeability of free space, 

θ ⇒ between → 

d l and → 

r . 

27. (i) On increasing distance between the coils, flux 

linked decreases, hence mutual inductance decreases 

µ 0 

N1N2A (ii) Q M = 

l 

∴ on increasing no. of turns, M increases 

(iii) when iron sheet (µr) is inserted 

µ 0 µ rN1N 

2A 

Then M = 

l 

∴ M increased, because for iron µr > > 1 

r 

) 

A

28 Parallel light rays incidenting over a lens converges 

to a point (convex lens) or seems to diverge from the 

point (concave lens) after refraction from lens. Then 

this point is called principle focus of lens. 

When lens is dipped into a liquid of refractive index 

greater than lens material refractive index then 

nature of lens gets changed. 

Focal length of lens is maximum for red and 

minimum for violet colour. 

I 

– µ1 = 1 

µ1 = 1 + 

µ2 

II 

R1 = + 20; R2 → ∞ ; 

1 ⎛ 1 1 ⎞ 

= (µ2 – 1) ⎜ − ⎟ 

f 

⎝ + 20 ∞ ⎠ 

1 ⎛ ⎞ 

= (2 –1) ⎜ − ⎟⎠ 

f ⎝ ∞ 

1 1 

20 

∴ f = + 20cm 

OR 

O 

µ1 

α 

i 

u P 

M 

R 

By snell's law 

m1 sin i = µ2 sin r 

r 

β γ 

v 

C 

µ2 

But for small aperture MP; 

µ1i = µ2r …(i) 

In ∆OCM; i = α + β 

and In ∆ICM; β = r + γ ⇒ r = β – γ. 

∴ By (i) 

µ1 (α + β) = µ2 (β – γ) 

⎡ MP MP ⎤ ⎡ MP MP ⎤ 

µ1 ⎢ + ⎥ = µ2 

⎣ OP PC 

⎢ − ⎥ 

⎦ ⎣ PC PI ⎦ 

⎡ 1 1 ⎤ ⎡ 1 1 ⎤ 

µ1 ⎢ + ⎥ = µ2 

⎣− 

u + R 

⎢ − ⎥ 

⎦ ⎣ + R + v ⎦ 

– 

µ 1 µ 1 µ 2 µ 2 

+ = − 

u R R v 

µ 2 µ 1 µ 2 − µ 1 

− = 

v u R 

I 

29. Principal of van deGraff Generator : (i) Let a 

small charged conducting shell of radius 

r be located inside a larger charged conducting shell 

of radius R. If they are connected with a conductor, 

then charge q from the small shell will move to the 

outer surface of bigger shell irrespection of its own 

charge Q. 

Here potential difference 


R 

r 

q 

= V(r) – V(R) 

q 

= 

4πε 

0 

⎟ ⎛ 1 1 ⎞ 

⎜ – 

⎝ r R ⎠ 

In this way, the potential of the outer shell increases 

considerably. 

(ii) Sharp pointed surfaces have larger charge 

densities, so these can be used to set up discharging 

action. 

Conducting 

S 

Shell 

Grounded 

Steal 

Tank 

C1 , C2 

metal 

Comb 

H 

vR 

C2 

Q 

Target 

Insulating 

Column 

Working :Let spray comb C1 be charged to a high +ve 

potential which spray +ve charge to the belt which in 

turn becomes positively charged. Since belt is 

moving up, so it carries this positive charge upward. 

Opposite charge appears on the teeth of collecting 

comb C2 by induction from the belt. As a result of 

this, positive charge appears on the outer surface of 

shell S. As the belt is 

moving continuously, so the charge on the shell S 

increase continuously. Consequently, the potential 

of the shell (S) rises, to a very high value. 

Now the charged particles at the top of the tub (T) 

are very high potential with respect to the lower end 

of the tube which is earthed. Thus these particles get 

accelerated downward and hit the target emerging 

from the tube.

Use : It can be used to accelerate particles which are 

used in nuclear physics for collision experiments. 

Or 

(i) Intensity of the Electric field at a point on the 

Axis of a Diople : 

A 

l l 

B E2 P E1 

–q O +q 

r 

The intensities E1 and E2 are along the same line in 

opposite directions. Therefore, the resultant intensity 

E at the point P will be equal to their difference and 

in the direction BP (since E1 > E2). That is, 

E = E1 – E2 

1 q 1 q 

= 

= 

= 

= 

4πε 

K 

0 

q 

4πε 

K 

0 

2 

( r – l) 

⎡ 1 

⎢ 

⎣( 

r – l) 

2 

– 

– 

4πε 

K 

( r 

1 

0 

+ 2 

l) 

q 

4πε0K 

⎥ ⎥ ⎡ 2 2 

( r + l) 

– ( r – l) 

⎤ 

⎢ 2 2 2 

⎢⎣ 

( r – l ) ⎦ 

q 

4πε 

K 

0 

⎡ 4lr 

⎤ 

⎢ 2 2 2 ⎥ 

⎣( 

r – l ) ⎦ 

⎤ 

⎥ 

⎦ 

( r 

+ l) 

1 ⎡ 2( 

2q 

l) 

r ⎤ 

= ⎢ 2 2 2 ⎥ 

4πε0K 

⎣( 

r – l ) ⎦ 

But 2ql = p (electric dipole moment). 

1 2p 

r 

∴ E = 

2 2 2 

4πε0K 

( r – l ) 

If l is very small compared to r (l

30. 

φ 

E 

Or 

Let a source of alternating e.m.f. be connected to a 

circuit containing a pure inductance only, Fig. 

Suppose the alternating e.m.f. supplied is 

represented by 

E = E0 

sin ωt 

L 

If dI/dt is the rate of change of current through L at 

any instant, then induced e.m.f. in the inductor at the 

same instant is = – L dI/dt . The negative sign 

indicated that induced e.m.f. opposes the change of 

current. To maintain the flow of current, the applied 

voltage must be equal and opposite to the induced 

voltage 

⎛ dI ⎞ 

i.e. E = – ⎜ – L ⎟ = Eo sin ωt 

⎝ dt ⎠ 

E0 or dI = sin ω t dt 

L 

Integrating both sides, we get 

E0 I = 

L 

E0 = 

L 

∫sin ωt dt 

A 

⎛ cosωt 

⎞ E0 ⎜ – ⎟ = – cos ωt 

⎝ ω ⎠ ωL E0 = – 

ωL sin ⎟ ⎛ π ⎞ 

⎜ – ωt 

⎝ 2 ⎠ 

E0 or I = sin (ωt – π/2) 

ωL …(i) 

The current will be maximum i.e. I = I0, when sin 

(ωt– π/2) = maximum = 1. 

E0 From (i), I0 = × 1 

ω L 

Putting in (i), we get 

I = I0 

sin( ωt 

– π/ 

2) 

…(ii) 

This is the form of alternating current developed. 

t 

CHEMISTRY 


1 

Ammonical AgNO3 

CH3CH2CH2CHO ⎯ ⎯⎯⎯⎯⎯⎯→CH3CH2CH2COOH Butanal 

( Tollens' 

reagent) 

Butanoic 

acid 

2. 3-Amino-2-chloro butanamide 

3. These are partly produced in body. 

Ex. Histidine and arginine 

4. yes, because O2 is paramagnetic 

5. In B.C.C no. of effective atoms are = 2 

In F.C.C no. of effective atoms are = 4 

6. In lyophillic colloids affinity is present between 

dispersed phase and dispersion medium and 

therefore it is more stable. 

7. r = k [NO] 2 [H2] 

8. 6-Amino 3-chloro 4-hydroxy hexanamide 

9. (i) Aldol condensation : Two molecules of an 

aldehyde or a ketone having at least one α-hydrogen 

atom, condense in the presence of a dilute alkali to 

give β-hydroxy aldehyde or β-hydroxy ketone. 

O 

OH 

CH3–C + HCH2CHO 

Dil. NaOH CH3–C–CH2CHO 

H 

H 

Ethanal Ethanal Aldol 

(ii) Gabriel phthalimide synthesis : Phthalimide on 

treatment with ethanoic KOH gives potassium 

phthalimide which on heating with a suitable alkyl halide 

gives N-substituted phthalimides. These upon hydrolysis 

with dil HCl under pressure give primary amines. 

CO 

CO 

Phthalimide 

NH + KOH (alc) 

–H2O 

C2H5I, ∆ 

–KI 

CO 

N–C2H5 

CO 

N-Ethylphthalimide 

CO 

N 

CO 

– H + 

Pot. Phthalimide 

H + / –H2O 

COOH 

C2H5NH2 + 

COOH 

Ethylamide 

Phthalic acid

10. 

(i) 

(ii) 

Benzene 

H 

CH3 – C = O 

Ethanal 

CH3COCl + Anhyd. AlCl3 

F.C.acylation 

CH3 – C = O + H2 

CH3 

Propanone 

CH3MgBr 

Dry ether 

Cu 

573 K 

COCH3 

Acetophenone 

H 

CH3 – C – OMgBr 

H + /H2O 

CH3 

H 

CH3 – C – OH 

CH3 

11. The compound is CH3CH2CHO, Its IUPAC name is 

propanal. The isomer is CH3COCH3, acetone. 

12. (i) Treat the compound with Lucas reagent (conc. 

HCl + anhy. ZnCl2) 2-propanol gives turbidity in 5 

min whereas ethanol gives no turbidity at room 

temperature 

ZnCl 

CH3CH2OH + HCl ⎯⎯⎯→ 

2 No reaction 

ZnCl2 

CH3CHCH3 + HCl 

OH 

CH3–CH–CH3 + H2O 

Cl 

Turbidity appears 

in 5 min 

(ii) Acetaldehyde reduces Tollen's reagent to 

silver mirror but acetone does not. 

CH3CHO + 2[Ag(NH3)2] + + OH – → 

CH3COO + 2H2O + 2Ag↓ + 4NH3 

Tollen's 

reagent 

CH3COCH3 ⎯⎯⎯⎯⎯⎯→ 

No action 

Sol.3 (a) Where concentration of reactant become unity or 

one litre rate of reaction is called as rate constant. 

(b) The reaction which has order greater than one 

but follows the kinetics of I st order reaction is called 

pseudo unimolecular reaction. 

CH3 COOC2H5 + H2O(excesss) ⎯⎯→ HCl 

CH3COOH + C2H5OH 

y = K[CH3COOC2H5] 

velocit 

14. (a) When NaCl is added neutralization of charge 

takes place which results in precipitation of 

colloidal solution and coagulation takes place. 

(b) Due to scattering of blue wavelength of light it 

makes image in our eye and sky appears blue. 

15. (i) The reaction involved is 

2H2O ⎯→ O2 + 4H + + 4e – 

2 mol H2O required ––– 4F charge 

∴ 1 mol H2O required ––– 2F charge 

(ii) 2FeO + H2O ––– Fe2O3 + 2n + + 2e – 

2 mol –––––––––––––2F 

1 mol –––––––––––––1F 

16. (i) Fluorine (ii) Iodine 

17. (i) Transition metals have unpaired e – 

(ii) Transition metals and compounds have 

unpaired e – 


18. 

19. 

(i) 

H 

| 

H – C = O + CH3CH2MgBr 

Methanal Grignard 

reagent 

(ii) 

H2O/H + 

H 

Dry ether 

H – C – OMgBr 

CH3CH2CH2OH + Mg 

Propan-1-ol 

CH2OH 

Benzyl alcohol 

CH2CH3 

KMnO4/H + 

OH 

Br 

COOH 

Benzoic alcohol 

(i) 2CH3 – CHI + 2Na 

| 

Dry ether 

CH3 CH3 – CH – CH – CH3 + 2NaI 

| | 

CH3 CH3 

2,3-Dimethyl butane 

(ii) (CH3)2CO 

LiAIH4 

CH3–CH–CH3 

| 

OH 

2-Propanol

(iii) 

OH 

COOH 

+ 

OCOCH3 

COOH 

Aspirin 

CH3CO 

CH3CO 

O 

+ CH3COOH 

20. (a) ––CH2–CH=CH2––CH––CH––CH2–– 

H 

H 

(b) ––CH2–C–– 

H 

C6H5 

(c) ––CH2–CH–– 

H 

CN 

C6H5 

21. Antibiotics are natural chemicals which kill/stop the 

growth 

of bacteria 

Types – 

(A) Bactericidal : Ex- penicillin, afloxacin 

(B) Bacteriostatic: Ex tetracycline, chloremphenicol 

22. Vitamins are organic compounds required in small 

quantity for normal functioning of body 

vitamin (C) is ascorbic acid. Deficiency- scurvy 

vitamin B1 is thiamine acid. Deficiency – beri-beri 

vitamin (B12) is cyanocobalamine. Deficiency anaemia 

23 (i) Rate of Reaction Rate constant of Reaction 

(ii) 

1. It is the speed at 

which the reactants 

are converted into 

the products at any 

moment of time 

2. It depends upon the 

concentration of 

reactants species at 

that 

time. 

moment of 

3. It generally 

decreases with 

progress of the 

reaction 

1. It is the constant of 

proportionality in the rate 

law equation 

2. It refers to the rate of the 

reaction at the 

specific point when 

concentration of every 

reacting species is unity 

3. It is constant and does not 

depend on the progress of 

the reaction 

Molecularity Order 

1. It is number of 

reacting species 

1. It is the sum of 

powers of 

undergoing 

simultaneous 

collisions in the 

reaction. 

2. It is a theoretical 

concept. 

3. It can have integral 

value only. 

4. It cannot be zero. 

5. It does not tell us 

anything about the 

mechanism of the 

reaction. 

concentration terms 

in the rate law 

expression 

2. It is determined 

experimentally. 

3. It can have even 

zero value. 

4. It can also have 

fractional values. 

5. It tells us about the 

slowest step in the 

mechanism and hence 

gives some clue about 

mechanism of the 

reaction. 

24. (a) The conductivity of all the ions of the solution 

which is kept between electrodes 1 cm apart and area 

of the electrodes 1 cm 2 . 

Molar conductivity can be defined as conductance of 

all the ions present in the solution which contain 1 

mol electrolyte in certain volume and solution is 

kept between electrodes 1 cm apart and area of the 

electrodes such that whole solution is present 

between them. 

(b) In this cell Zn acts as anode and Ag as cathode 

° 

E cell = 

° 

E Ag – 


° 

E Zn 

= 0.344 – (– 0.76) = 1.104 V 

∆G° = – n F E°cell = – 2 × 96500 × 1.104 

= – 2.13 × 10 5 J 

zM 

25. (A) We know d = 

V × N 

⇒ 7.86 = 

( 2. 

68× 

10 

A 

Z× 

56 

– 8 

z ~ – 2 i.e. B.C.C. structure 

) 

3 

23 

× 6. 

02× 

10 

(B) Total no. of atoms surrounding a particular atom 

in crystal structure is called coordination number. 

(i) In C.C.P → C.No. 12 ; 

(ii) In B.C.C → C.No. 8 

26. (a) 27Co +3 = [Ar] 

Octahedral geometry 

3d 4s 4p 

d 2 sp 3

NH3 

NH3 

NH3 

Co +3 

NH3 

NH3 

NH3 

No unpaired e – therefore diamagnetic 

(b) 28Ni +2 = [Ar] 

CN – 

CN – 

Ni +2 

3d 4s 4p 

CN – 

CN – 

Square planar 

dsp 2 

No unpaired e – ∴ Diamagnetic in nature 

3d 4s 4p 

(c) 28Ni = [Ar] 

CO 

CO 

Ni 

CO 

CO 

sp 3 

Tetrahedral geometry 

No unpaired e – ∴ Diamagnetic in nature 

27. (a) (i) K2Cr2O7 + 7H2SO4 + 6KI 

⎯→ Cr2(SO4)3 + K2SO4 + 7H2O + 3I2 

(ii) 6 KMnO4 + 6 KOH + KI 

⎯→ 6K2MnO4 + 3H2O + KIO3 

(b) Lanthanoids have almost similar properties. 

28. (a) According to Henry's law mole fraction of a gas 

is directly proportional to the pressure at which gas 

is disolved 

p = kHx 

App. (1) Dissolution of gases in cold drinks 

(2) Dissolution of O2 in haemoglobin 

(b) π = CRT 


8 

= 

180 

1000 

× × 0.0821 × 298 = 2.446 atm 

10 

∆Tf = Kfm = 1.86 × 0.1 = 0.186 

T = – 0.186ºC 

29. (a) Acetylene is first oxidized with 40% H2SO4 in 

the presence of HgSO4 

40%, 

H SO 

2 4 

H–C≡C – H + H2O ⎯⎯⎯⎯⎯→CH3–CHO 1% 

HgSO 

4 

Acetylene Acetaldehyde 

Acetaldehyde is finally oxidized to acid with air in 

the presence of manganous acetate catalyst 

Manganous 

CH3CHO + [ O] 

⎯ ⎯⎯ 

⎯ 

Acetaldehyde 

Acetate 

⎯ → 

Ca( 

OH) 

CH3COOH 

Acetic acid 

2 

(b) (i) CH3COOH ⎯⎯⎯⎯→ 

(CH3COO)2 Ca 

∆ 

⎯⎯ 

Ca 

⎯⎯⎯→ 

(HCOO) 2 

Ca( 

OH) 

Calcium acetate 

CH 3CHO 

+ 2CaCO3 

Acetaldehyde 

2 

(ii) CH3COOH ⎯⎯⎯⎯→ 

(CH3COO)2Ca 

Acetic acid 

⎯⎯→ CH 3COCH3 

+ CaCO3 

Acetone 

∆ 

(c) When heated with I2 + NaCO3 Solution, acetone 

gives yellow crystals of iodoform CH3COCH3 + 

3NaOI → CH3I + CH3COONa 

Acetone Yellow ppt. 

(Iodoform) 

Acetic acid does not give iodoform test. 

(d) The carbonyl group in – COOH is inert and does 

not show nucleophilic addition reaction like 

carbonyl compounds. It is due to resonance 

stabilization of carboxylate ions. 

R – C = O 


O – 

R – C = O – 

Or 

(a) (i) Due to smaller + I-effect of one alkyl group in 

aldehydes as a compared to larger +I-effect of two 

alkyl groups, the magnitude of positive charge on the 

carbonyl carbon is more in aldehydes than in 

ketones. As a result nucleophilic addition reaction 

occur more readily in aldehyde than in ketones. 

(ii) The boiling points of aldehydes and ketones are 

lower than corresponding acids and alcohols due to 

absence of intermolecular hydrogen bonding . 

(iii) Aldehydes and ketones undergo a number of 

addition reactions as both possess the carbonyl 

functional group which reacts a number of 

nucleophiles such as HNC, NaHSO3, alcohols, 

ammonia derivatives and Grignard reagents. 

(b) (i) Distinction between acetaldehyde and 

benzaldehyde: Acetaldehyde and benzaldehyde can 

be distinguish by Fehling solution. 

Acetaldehyde give red coloured precipitate with 

Fehling solution while benzaldehyde does not. 

2+ 

− 

CH 

3CHO 

+ 2Cu 1442+ 

45 OH 43⎯⎯→ 

Fehling Solution 

O

CH COO Cu O H2O 

red ppt. 

2 

3 

− 

+ + 

(ii) Distinction between Propanone and Propanol : 

Propanone (CH3COCH3) and propanol 

(CH3CH2CH2OH) can be distinguish by iodoform 

test. Propanone when warmed with sodium 

hypoiodide (NaOI) i.e. I2 in NaOH, it gives yellow 

ppt of idoform while propanol does not respond to 

iodoform test. 

CH 3 COCH 3 

Pr opanone 

+ 3NaOI 

⎯⎯→ 

CH3 3 

Yellow ppt 

I ↓ + CH COONa + 2NaOH 

2× 

M 

30 (a) Mass of unit cell = 

23 

6. 

023× 

10 

Density = 7.2 g cm –3 

α = 288 pm = 288 × 10 –10 cm 

Mass 

Now density = 

Volume 

∴ 7.2 = 

6. 

023× 

10 

2× 

M 

23 

× ( 288× 

10 

– 10 

7. 

2× 

6. 

023× 

10 × ( 288) 

× 10 

or M = 

2 

= 51.79 g mol –1 . 

23 

3 

) 

3 

– 30 

(b) Those compounds containing two or more 

halogen atoms in their molecules are known as 

Interhalogen compounds. Properties : 

(i) They are covalent compounds and diamagnetic in 

nature. 

(ii) They are more reactive than the constituent 

halogens. It is because A – X bond is relatively 

weaker than X – X bond. 

(iii) They are very good oxidizing agents. 

Their melting point and boiling point are higher than 

halogens and increases with increase in the 

difference of electronegativity. 

MATHEMATICS 

Section A 

x + 2 y + 3 z − 3 

1. Given line is = = . 

− 1 7 3 / 2 

∴ Its direction ratios are –1, 7, 3/2 

Or –2, 14, 3 

Equation of line pasing through (1, 2, 3) is 

x + 2 y − 2 z − 3 

= = . 

− 2 14 3 


2. 

→ 

b + → 

c = î + 3 jˆ + kˆ + î + kˆ = 2 î + 3 jˆ + 2 kˆ Projection of → 

b + → 

c on → ( b + c ). a 

a = . 

| a | 

( 2iˆ 

+ 3 ˆj 

+ 2kˆ 

).( iˆ 

+ 2 ˆj 

+ kˆ 

) 2 + 6 + 2 10 

= 

= = 

2 2 2 

( 1) 

+ ( 2) 

+ ( 1) 

1+ 

4 + 1 6 

3. The vectors → 

a , → 

b , → 

c are coplanar if → 

a . [ → 

b × → 

c ] = 0 

– 4 

– 1 

– 8 

– 6 

4 

– 1 

– 2 

3 

λ 

= 0 

– 4 + (4λ + 3) + 6 (– λ + 24) – 2 (1 + 32) = 0 

–16λ – 12 – 6λ + 144 – 66 = 0 

– 22λ + 66 = 0 

– 22λ = –66 ⇒ λ = 3 

4. order = 2 

Q The equation can't be expressed in polynomial 

from 

⇒ degree is not defined 

5. 

4 

4 

2 

(sin x + cos x) 


(sin x − cos x) 

⇒ ∫ dx 

2 2 2 2 2 


− 2sin 

xcos 

x 

4 

4 


( 1) 

(– cos 2x) 

⇒ ∫ dx 

4 4 

sin x + cos x 

= – 

sin 2x 

+ c 

2 

⎡2 

3⎤ 

⎡2 

4⎤ 

6. A = ⎢ ⎥ A′ = 

⎣4 

5 

⎢ ⎥ 

⎦ ⎣3 

5⎦ 

→ 

⎡0 A – A′ = ⎢ 

⎣1 

−1⎤ 

0 

⎥ 

⎦ 

⎡0 (A – A′)′ = ⎢ 

⎣1 

−1⎤ 

′ ⎡ 0 

0 

⎥ = ⎢ 

⎦ ⎣−1 

1⎤ 

0 

⎥ 

⎦ 

⎡0 −1⎤ 

⇒ (A – A′) = – ⎢ ⎥ 

⎣1 

0 ⎦ 

⇒ (A – A′)′ = – (A – A′) 

⇒ (A – A′) is skew symmetric matrix. 

2 

→ 

→ 

2 

2

7. f(A) = A 2 – 5A + 7 I2 

⎡ 3 1⎤ 

⎡ 3 1⎤ 

⎡ 3 1⎤ 

⎡1 

0⎤ 

= ⎢ ⎥ . 

⎣−1 

2 

⎢ ⎥ – 5 

⎦ ⎣−1 

2 

⎢ ⎥ + 7. 

⎦ ⎣−1 

2 

⎢ ⎥ 

⎦ ⎣0 

1⎦ 

⎡ 8 5⎤ 

⎡15 

5 ⎤ ⎡7 

0⎤ 

= ⎢ ⎥ – 

⎣− 

5 3 

⎢ ⎥ + 

⎦ ⎣− 

5 10 

⎢ ⎥ 

⎦ ⎣0 

7⎦ 

⎡0 

0⎤ 

= ⎢ ⎥ = Null Matrix. 

⎣0 

0⎦ 

8. Area of ∆ = 0 

1 

⇒ 

2 

k 

− k + 1 

− 4 − k 

2 − 2k 

2k 

6 − 2k 

1 

1 

1 

= 0 

k . [2k – 6 + 2k] – (2 – 2k) [–k + 1 + 4 + k] 

+ 1 [(–k + 1) (6 – 2k) –2k (–4 – k)] = 0 

k [4k – 6] – (2 – 2k) (5) + [2k 2 – 8k + 6 + 8k + 2k 2 ] = 0 

4k 2 – 6k – 10 + 10k + 4k 2 + 6 = 0 

8k 2 + 4k – 4 = 0 

2k 2 + k – 1 = 0 

(2k – 1) (k + 1) = 0 ⇒ k = 1/2, – 1 

9. y = f (e x ) 

y' = f '(e x ). e x 

x 

x 

y " = f ' ( e ). e + e . f " ( e ). e 

10. f (g(x)) = g ( x) 

= x 1 

x 

2 − 

g(f (x)) = f (x) 2 – 1 = ( x ) 1 = x – 1 

x 

2 − 

Section B 

11. A : 3 cards have the same number 

⎛ 4 ⎞ 

n(A) = 13 C(4, 3) = 13 ⎜ ⎟ 

⎜ ⎟ 

= 13 (4) = 52 

⎝ 3 1⎠ 

52 

n(S) = C(52, 3) = = 22100 

3 49 

Required probability 

n( 

A) 

52 13 1 

= P(A) = = = = 

n( 

S) 

22100 5525 425 

OR 

Let E : Candidate Reaches late 

A1 = Candidate travels by bus 

A2 : Candidate travels by scooter 

A3 : Candidate travels by other modes of 

transport 

x 

3 1 3 

P(A1) = , P(A2) = , P(A3) = 

10 10 5 

1 1 

P(E/A1) = , P(E/A2) = , P(E/A3) = 0 

4 

3 

∴ By Baye's Theorem 

P(A1/E) = 

) P( 

E / A ) + 

) P( 

E / A1) 

) P( 

E / A ) + 

) P( 

E / A 


= 

P( 

A 

1 

3 1 

× 

10 4 

3 1 

+ + 0 

40 30 

→ 

1 

9 

= 

13 

P( 

A 

P( 

A 

1 

2 

2 

P( 

A 

12. Here, A = 3 iˆ 

+ 2 ˆj 

+ 9kˆ 

; B = iˆ 

+ λˆj 

+ 3kˆ 

+ → → 

⇒ a b = ( 3iˆ 

+ 2 ˆj 

+ 9kˆ 

) + ( iˆ 

+ ˆj 

+ kˆ 

) 

= 4 iˆ 

+ ( 2 + λ) 

ˆj 

+ 12 kˆ 

− → → 

a b = ( 3iˆ 

+ 2 ˆj 

+ 9kˆ 

) − ( iˆ 

+ λˆj 

+ 3kˆ 

) 

= 2 iˆ 

+ ( 2 − λ) 

ˆj 

+ 6kˆ 

⎡ → →⎤ 

⎡→ →⎤ 

Since ⎢a 

+ b ⎥ ⊥ ⎢a 

− b ⎥ we have 

⎣ ⎦ ⎣ ⎦ 

→ 

⎡ → →⎤ 

⎡→ →⎤ 

⎢a 

+ b ⎥ ⋅ ⎢a 

− b ⎥ = 0 

⎣ ⎦ ⎣ ⎦ 

⇒ [ 4i 

ˆ + ( 2 + λ) 

ˆj 

+ 12kˆ 

] [ 2iˆ 

+ ( 2 − λ) 

ˆj 

+ 6kˆ 

] = 0 

⇒ 4 × + (2 + λ) × (2 – λ) + 12 × 6 = 0 

⇒ 8 + 4 – λ 2 + 72 = 0 

⇒ λ 2 = 84 ⇒ λ = ± 2 21 

13. Equation of plane passing through the inter sections 

of planes 

x + 2y + 32 – 4 = 0 and 2x + 4 – z + 5 = 0 

(x + 2y + 32 – 4) + λ (2x y – z + 5) = 0 …(i) 

x + 2y + 32 – 4 + 2λx + λy –λz + 5λ = 0 

(1 + 2λ) x + (2 + λ) y + (3 –λ) z – 4 + 5λ = 0 

Since the plane (i) is perpendicular to 

5x + 3y + 6z + 8 = 0 

∴ (1 + 2λ) ⋅ 5+ (2 + λ) ⋅ 3 + ( 3 – λ) ⋅ 6 = 0 

5 + 10λ + 6 + 3λ + 18 – 6 λ = 0 

7λ + 29 = 0 ⇒ 

−29 

λ = 

7 

∴ required equation of plane is 

−29 

(x + 2y + 3z – 4) (2x + y – z + 5) = 0 

7 

7x + 14 y + 21z – 28 – 54x – 29y + 29z – 145 = 0 

– 47 x – 15y + 50z –173 = 0 

47x + 15y – 50z + 173 = 0 

3 

3 

)

OR 

The equation of a plane passing through the 

intersection of the given planes is 

(4x – y + z –10) + λ(x + y – z –4) = 0 

⇒ x(4 + λ) + y (λ –1) + z (1 –λ) –10 – 4λ = 0 

This plane is parallel to the line with direction ratios 

proportional to 2, 1,1 

∴ 2(4 + λ) + 1(λ –1) + 1(1 –λ) = 0 ⇒ λ = – 4 

Putting λ = – 4 in (i), we obtain 

5y – 5z – 6 = 0 

This is the equation of the required plane. 

Now, length of the perpendicular from (1, 1, 1) on (ii) 

is given by 

d = 

5× 

1− 

5× 

1− 

6 

5 

2 

+ ( −5) 

sin( x − a + a) 

14. ∫ dx 

sin( x − a) 

2 

3 2 

= 

5 

sin( x − a) 

cos a + cos( x − a) 

sin a 

∫ dx 

sin( x − a) 

= cos a. x + sin a. ln sin (x – a) + c 

15. sin x = t 

cos x dx = dt 

dt 

∫ 2 

t − 2t 

− 3 

dt 

∫ (t −1) 

− 

2 2 

2 

log | ( t − 1) 

+ ( t −1) 

− 2 | + c replace t 

16. take log 

m log x + n log y = (m + n) log (x + y) 

m n dy ( m + n) 

⎛ dy ⎞ 

diff. + = ⎜1+ 

⎟ 

x y dx x + y ⎝ dx ⎠ 

nx − my dy nx − my dy 

⇒ 

= ⇒ = y / x 

y( 

x + y) 

dx ( x + y) 

x dx 

17. C1 → C1 + C2 + C3 

(a + b + c) 

1 

1 

1 

3b 

2 

− a + b 

− c + b 

2 

− a + c 

− b + c 

3c 

R2 → R2 – R1, R3 – R1 

1 − a + b − a + c 

(a + b + c) 

1 

1 

2b 

+ a 

− c + a 

− b + a 

2c 

+ a 

(a + b + c) [(2b + a) (2c + a) – (a – b) (a – c)] 

= (a + b + c) [3bc + 3ab + 3ac] 

= 3 (a + b + c) (ab + bc ca) hence proved. 

x = 0 ⎤ 

18. at y-axis 

y = b 

⎥ ⇒ pt = (0, b) 

⎦ 


Now 

dy b −x 

/ a 

= − e 

dx a 

⎛ dy ⎞ b 

⎜ ⎟ = − 

⎝ dx ⎠ a 

0, 

b 

b 

equation of tangent y – b = – (x – 0) 

a 

x 

⇒ + y / b = 1 

a 

19. centre = (h, 0) rad. = h equation is 

(h, 0) 

(x – h) 2 + (y – 0) 2 = h 2 

⇒ x 2 + y 2 – 2hx = 0 … (1) 

Deff. 2x + 2y . y 1 – 2h = 0 ⇒ h = x + y y' 

…(2) 

From (1), (2) 

2 

2 

x + y = 2( 

x + yy' 

) x 

OR 

Let P(x, y) be any point on the curve. The equation of 

the normal at P (x, y) to the given curve is 

1 

Y – y = – (X –x) … (i) 

dy / dx 

It is given that the normal at each point passes through 

(2, 0). Therefore, (i) also passes through (2, 0). Putting 

Y = 0 and x = 2 in (i), we get 

1 

0 – y = – (2 –x) 

dy / dx 

dy 

⇒ y = 2 – x 

dx 

⇒ ydy = (2 – x) dx [On integrating both sides] 

2 

2 

y 

⇒ 

2 

( 2 − x) 

= – 

2 

+ C 

⇒ y 2 = – (2 – x) 2 + 2C … (ii) 

This passes through (2, 3). Therefore, 

9 

9 = 0 + 2C ⇒ C = 

2 

9 

Putting C = in (ii), we get 

2

y 2 = – (2 – x) 2 + 9 

This is the equation of required curve. 

20. Q f (x) is conti. at x = 1 

∴ f (1) = f (1–) 

⇒ 1 = 

lim 

h→0 

a (1 – h) 2 – b 

⇒ 1 = a – b …(1) 

x 

−x 

y 10 −10 

21. Let = 

1 x −x 

10 + 10 

Componendo – dividendo rule 

⇒ 

⇒ 

x 

−x 

−x 

y −1 

( 10 −10 

) − ( 10 + 10 

= 

x −x 

x − 

y + 1 ( 10 −10 

) + ( 10 + 10 

−x 

y −1 

2. 

10 

= − 

y + 1 2. 

10 

y + 1 2x 

⇒ = 10 

1− 

y 

1 

⇒ x = ln10 

2 ⎟ ⎛ y + 1 ⎞ 

⎜ 

⎝1 

− y ⎠ 

−1 

1 ⎛ x + 1⎞ 

∴ f ( x) 

= l n10⎜ 

⎟ 

2 ⎝1 

− x ⎠ 

1 + 1 

f(1) = = 1 

2 

x 

OR 

2 

2 

f(2) = = 1 

2 

many-one function 

If n → odd natural number then 2n –1 is also odd 

number 

2 n −1 

+ 1 

f(2n –1) = = n 

2 

If n → even natural number then 2n is also an even 

natural number 

2n 

f(2n) = = n 

2 

⇒ f is onto function. 

22. Let y = sin –1 ⎜ ⎟ 

⎝ + ⎠ 

2 

1 x 

put x = tan θ 

⎛ 

⎛ 

2x 

2 tan θ 

y = sin –1 ⎜ 2 ⎟ 

⎝1 

+ tan θ ⎠ 

⎞ 

⎞ 

1 

x 

x 

) 

) 

1 

y = sin –1 (sin 2θ) 

y = 2tan –1 x 

dy 2 

= 

2 dx 1+ 

x 

z = tan –1 x 

dz 1 

= 

dx 2 

1+ 

x 


Now 

dy 2 / 1+ 

x 

= = 2 

2 

dz 1/ 

1+ 

x 

2 

Section C 

23. Let x denotes the number of spades. Clearly x can 

take the values 0, 1, 2, or 3. 

13 1 

Probability of getting a spade = = 

52 4 

1 3 

Probability of not getting a spade = 1 – = 

4 4 

3 3 3 27 

P(x = 0) = P (No spade) = × × = 

4 4 4 64 

P(x = 1) = P (One spade and two non spade) 

= 3 1 3 3 27 

C1 × × × = 

4 4 4 64 

P(x = 2) = P (Two spades and one non spade) 

= 3 1 1 3 9 

C2 × × × = 

4 4 4 64 

1 1 1 1 

P(x = 3) = P (Three spade) = × × = 

4 4 4 64 

∴ The probability distribution of the random variable x is 

x 0 1 2 3 

P(x) 

27 

64 

27 

64 

24. The required line is to pass through 

→ 

a = 2iˆ − 3 ˆj 

− 5kˆ 

and is ⊥ to the plane given by 

→ 

9 

64 

1 

64 

r .( 6iˆ 

− 3 ˆj 

+ 5kˆ 

) + 2 = 0 

……(i) 

→ 

Here the vector b = 6 iˆ 

+ 3 ˆj 

+ 5kˆ 

is normal to the 

plane (1) 

⇒ The required line is along the direction of this 

vector. Hence its equation is 

→ 

→ 

→ 

→ 

r = a + t b 

⇒ r = 2iˆ − 3 ˆj 

− 5kˆ 

+ t( 

6iˆ 

− 3 ˆj 

+ 5kˆ 

) 

Now line (2) meets the plane (1), when 

….(ii)

[ ( 2i 

ˆ 3 ˆj 

− 5kˆ 

) + t( 

6iˆ 

− 3 ˆj 

+ 5kˆ 

) ] 

− ⋅ ( 6i 

ˆ − 3 ˆj 

+ 5kˆ 

) + 2 = 0 

i.e. when 

6 (2 + 6t) + 3 (3 + 3t) + 5 (5t – 5) = –2 

1 

i.e. when t = 

35 

Substituting this value of t in (2), we get 

→ 

1 

r = 2iˆ − 3 ˆj 

− 5kˆ 

+ ( 6iˆ 

− 3 ˆj 

+ 5kˆ 

) 

35 

1 

= ( 76iˆ 

− 108 ˆj 

−170kˆ 

) 

35 

∴ The required point of intersection is 

⎡76 −108 

−170⎤ 

⎢ , , ⎥ 

⎣35 

35 35 ⎦ 

25. Let the distance covered at the speed 25 km/hour = x km 

and the distance covered at the speed 40 km hour 

= y km 

maximum distance z = x + y 

subject to constraints 

2x + 5y ≤ 100 

+ ≤ 1 

25 40 

y x 

x, y ≥ 0 

Table for 2x + 5y = 100 …(i) 

x 0 50 

y 20 0 

Table for + = 1 

25 40 

y x 

x 0 25 

y 40 0 

Plot lines 2x + 5y = 100 and + = 1 

25 40 

y x 

on a graph 

paper. The shaded region satisfy the given 

inequalities. 

D(25,0) B 

30 40 50 


50 

(0, 40) 40 

30 

A (0, 20) 

20 

10 

O 

Y 

C 

10 

20 

⎡50 

40⎤ 

P⎢ 

, ⎥ 

⎣ 3 3 ⎦ 

X 

I 

II 

The coordinates of feasible region is A (0, 20) 

⎡50 

40⎤ 

P ⎢ , ⎥ , D (25, 0) 

⎣ 3 3 ⎦ 

Corner points z = x + 5 

A (0, 20) z = 0 + 20 = 20 

⎡50 

40⎤ 

P ⎢ , ⎥ 

⎣ 3 3 ⎦ 

50 40 

z = + = 30 

3 3 

D (25, 0) z = 25 + 0 = 25 

∴ Max. distance covered = 30 km 50/3 km at the 

speed of 25 km/hour and 40/3 km at the speed of 

40 km/hour 

26. I st part : Area = [( 2x 

− x ) − ( −x)] 

dx 

area = 9/2 

(0, 0) 

∫ 

3 

0 

⎡3 

= ⎢ 

⎢⎣ 

2 

x 

2 nd part : 

Both curve intersect at 

( 2 3, 

2), 

( − 

2 3, 

2) 

2 

2 

3 ⎤ 

− ⎥ 

3 ⎥⎦ 

x 

3 

0 

(2, 0) 

(3, –3)

(0, 4) 

( 2 

3, 

2) 

(4, 0) 

⎡ 2 

4 

2 ⎤ 

Area = 2 ⎢ + − ⎥ 

⎣∫ 

6y dy 

0 ∫ 16 y dy 

2 ⎦ 

⎡ 

4 

2 

⎤ 

3 / 2 2 

= 

⎛ 1 

2 1 −1 

⎞ 

2⎢ 6. 

( y ) 0 + ⎜ y 16 − y + × 16 sin y / 4⎟ 

⎥ 

⎢⎣ 

3 ⎝ 2 

2 

⎠ 2 ⎥⎦ 

= 

4 3 16π 

+ 

3 3 

π/ 2 

27. I = ∫0 By property 

π/ 2 

I = ∫0 2I = ∫ π 

0 

2I = ∫ π 

log sin xdx 

…(1) 

log cos x dx 

…(2) 

/ 2 

/ 2 

⎛ sin 2x 

⎞ 

log ⎜ ⎟dx 

⎝ 2 ⎠ 

logsin 2xdx 

− (log2) 

∫ 

π 2 / 

dx 

0 

0 

π 

2I = I1 – log2 

2 

…(3) 

Now In I1 2x = t 

dx = 1/2 dt 

I1 = ∫ π 1 

logsin 

t dt 

2 0 

π 2 

= ∫ / 1 

. 2 logsin 

t dt 

2 0 

I1 = I 

…(4) 

From (3), (4), 2I = I – π/2 log 2 

I = – π/2 log 2 

28. System of equations can be written as : 

⎡1 

1 1 ⎤ ⎡x⎤ 

⎡5⎤ 

⎢ ⎥ 

⎢ 

2 1 −1 

⎥ 

. 

⎢ ⎥ 

⎢ 

y 

⎥ 

= 

⎢ ⎥ 

⎢ 

2 

⎥ 

⎢⎣ 

2 −1 

1 ⎥⎦ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

2⎥⎦ 

AX = B 

X = A –1 . B = ⎟ ⎛ Adj( 

A) 

⎞ 

⎜ B 

⎝ | A | ⎠ 

⎡ 0 − 2 − 2⎤ 

⎡5⎤ 

⎡x⎤ 

⎡1⎤ 

− 1 

X = 

⎢ 

⎥ 

8 ⎢ 

− 4 −1 

3 

⎢ ⎥ 

⎥ ⎢ 

2 

⎥ 

⇒ 

⎢ ⎥ 

⎢ 

y 

⎥ 

= 

⎢ ⎥ 

⎢ 

2 

⎥ 

⎢⎣ 

− 4 3 −1⎥⎦ 

⎢⎣ 

2⎥⎦ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

2⎥⎦ 

x = 1, y = 2, z = 3 

OR 

System of equations can be written as : 

⎡2 

−1 

2 ⎤ ⎡x⎤ 

⎡ 3 ⎤ 

⎢ ⎥ 

⎢ 

1 1 2 

⎥ 

. 

⎢ ⎥ 

⎢ 

y 

⎥ 

= 

⎢ ⎥ 

⎢ 

2 

⎥ 

⎢⎣ 

2 3 −1⎥⎦ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

− 2⎥⎦ 

A . X = B 

X = A –1 ⎛ Ad jA 

⎞ 

. B = ⎜ ⎟ 

⎜ ⎟ 

. B 

⎝ | A | ⎠ 

⎡− 

7 7 0 ⎤ ⎡ 3 ⎤ 

1 

X = 

⎢ 

⎥ 

⎢ 

5 − 6 − 2 

− 

⎥ 

. 

⎢ ⎥ 

7 

⎢ 

2 

⎥ 

⎢⎣ 

1 − 4 1 ⎥⎦ 

⎢⎣ 

− 2⎥⎦ 

⎡x⎤ 

⎡ 1 ⎤ 

⇒ X = 

⎢ ⎥ 

⎢ 

y 

⎥ 

= 

⎢ ⎥ 

⎢ 

−1 

⎥ 

⎢⎣ 

z⎥⎦ 

⎢⎣ 

1 ⎥⎦ 

x = –1, y = –1, z = 1 

29. Let length, width → x 

height → y 

volume v = x 2 y …(1) 

are s = x 2 + 4xy …(2) 

s = x 2 4 v 

+ 

x 

(from (1), (2)) 

ds 4v 

d s 8v 

= 2x – , = 2 + > 0 

dx 

2 2 3 

x dx x 

Now 

ds 

4 

= 0 ⇒ 2x – 0 

2 dx 

= 

v 

x 

⇒ x = 2y (Q v = x 2 y) 


2

MOCK TEST-3 (SOLUTION) 

PHYSICS 

1. There are many atoms in an atomic hydrogen gas 

sample and electrons of different atoms can make 

different transitions and different wavelength 

radiations are emitted. 

2. Frequency remains constant when light passes from 

one medium to other. 

3. Those sources which have constant phase difference 

are called coherent sources. These are derived from 

a single sources. 

4. Kinetic energy depends on frequency of incident 

radiations not on intensity so kinetic energy remains 

same. 

5. Average power over a complete cycle of ac through 

an ideal inductor (Resistance = 0) is zero. 

Pav = (Irms) × (Vrms) cos φ 

R 

cos φ = = 0 

Z 

Pav = 0 

6. A 

1A 2Ω 1Ω 

+ – 

B 

VA – VB = 1 × 2 + 2 + 1 × 1 = 5 V 

7. X–rays –→ 3 × 10 19 to 1 × 10 16 Hz 

Microwave –→ 3 × 10 11 to 1 × 10 9 Hz 

U-V rays –→ 5 × 10 17 to 8 × 10 14 Hz 

Radio Waves –→ 3 × 10 7 to 3 × 10 4 Hz 

8. φ = → → 

q 

E . A , ∆ φ = 

ε 

φi = EA cos 180° φf = EA cos 0° 

φi = – EA φf = + EA 

φi = – 5 × 10 5 V– m φf = 4 × 10 5 V– m 

– 5 × 10 5 + 4 × 10 5 = 

q 

0 

ε 0 

, q = – ε0 × 10 5 Cb 

9. Intensity of radiations is 64 = 2 6 times the safe value. 

So after six half lines it would be safe so 

t = 6T = 6 × 2 = 12 hours. 

MOCK TEST– 3 PUBLISHED IN SAME ISSUE 

10. On temperature rise more valance band electrons 

become free and more free electrons and hole pair 

will be produced and conductivity of semiconductor 

increases. 

11. In optical fibres refractive index of cladding is less 

than core. So that electromagnetic wave traveling in 

core when incidents other cladding at an angle 

greater than ic. Total internal reflection will occur 

which is necessary for transmission of signals from 

one place to other. 

12. Object should be placed at 2F of convex lens so as to 

form image of same size. It can't happen in case of 

concave lens because it always forms a diminished 

image. 

13. Using d = 2 Rh T + 2Rh 

R 

= R ( h h ) 

2 T R + 

= 20.8 5 km = 46.5 km 

14. When a charge +Q is given to the insulated plate, 

then a charge –Q is induced on the nearer face of the 

plate. 

+ Q – Q 


E 

Net electric field between the plate 

σ V 

E = also, E = 

ε0 

d 

∴ V = Ed = 

with dielectric. C' = 

σ d Q ε0A 

= ⇒ C = 

ε C d 

0 

ε0ε 

A 

r 

d

15. X = ? 

Y = 20 Ω 

l1 = 40 cm 

l – l1 = 60 cm 

X 

B 

Y 

16. 

A 

l1 

V1 

D 

V1 ∝ l1 

V2 ∝ (l – l1) 

X Y 

l ( l – l 

= 

1 

1 

) 

G 

l – l1 

V2 

X 20 X 1 

= or = or 

40 60 40 3 

40 

X = Ω 

3 

When X and Y are interchanged 

Y B X 

2 

A 

D 

G 

l2 l – l2 

Y 

l = 

X 20 40 

⇒ = 

( l – l ) l 3 ( 100 – l 

2 

1 2 

= 

l 2 3( 

100 – l 2) 

300 – 3l2 = 2l2 or 5l2 = 300 

l2 = 60 cm 

λ1λ 

2 F = 

2πε 

d 

0 

17. (i) Vcom = 

(ii) Ui = 

Uf = 2 

1 

λ1 

2 

C1V1 

+ C2V2 

C + C 

1 

C 

2 

2 

1V1 

+ 

2 

1 

C 

2 

1 

(C1 + C2) V 

d 

2 

V 

2 

com 

2 

2 

λ2 

2 

C 

) 

C 

E0 18. B0 = 

c 

19. Work-function : Minimum energy needed by a free 

electron to remove it from metallic surface. 

Threshold frequency : Minimum frequency 

required for photo-electron emission. 

Threshold wavelength : Maximum wavelength of 

incident radiations required for photoelectric 

emission. 

Stopping potential : Minimum potential required 

across the photo cell to completely stop the photo 

current. 

20. (En) = 

– 13. 

6 

eV, for ground state n = 1 

2 

n 

∴ (E1) = – 13.6 eV 

For hydrogen like atom, (En) HLA = Z 2 (En) H 


2 × 

Z 13. 

6 

= eV 

2 

n 

For He + atom, Z = 2 and n = 2 

⎡ – 13. 

6⎤ 

∴ (E2)He = 4 ⎢ ⎥ = – 13.6 eV 

⎣ 4 ⎦ 

5Dλ −2 

21. = 1. 

5× 

10 m 

d 

d = 0.56 × 10 –3 m 

D = 2.8 m 

5× 2. 

8× 

λ 

− 

∴ 

= 1. 

5× 

10 

−3 

0. 

56× 

10 

∴ λ = 6 × 10 –7 m 

∴ λ = 6000 Å 

c 

22. Source frequency ν = 

λ 

23. 

x 

Band width of system = ν = 

100 

2 

× 

λ 

c x 

100 

Number of channels N which can be transmitted 

simultaneously can be found out by dividing band 

width by the system with band width of one channel. 

i.e., 

R R 

= 

R R 

1 

2 

3 

4 

N = 

⎛ xc ⎞ 

⎜ ⎟ 

⎝100× 

λ ⎠ xc 

= 

F 100λF 

24. (1) Resistance = R 

(2) Reactance are of two types 

(a) Capacitive Reactance 

XC = C 

1 

ω

(b) Inductive Reactance 

XL = ωL 

(3) Impendence = Z = 

(4) 

XC 

ν 

2 

R + ( X 

XL 

L 

– X 

25. Force = F = Bil 

For current (i) 

2 2 

⎛ emf ⎞ Bvl B vl 

i = ⎜ ⎟ = ⇒ F = 

⎝ R ⎠ R 

R 

26. Cyclotron work on the fact that a positively charged 

particle can be accelerated to a sufficiently high 

energy with the help of smaller values of oscillating 

electric field by making it to cross the same electric 

field time and again with use of strong magnetic 

field. 

North 

Magnet 

H.F 

oscillator 

Dees D2 

27. 

∫ r 

qin 

E. 

ds = 

∈ 

E.2πrl = 

E = 

λ 

2π ∈0 

0 

λ· l 

∈ 

0 

r 

South 

B (Magnetic Field) 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

λ 

r 

Dees D1 

l 

C 

) 

ν 

2 

target 

28. Bending of light from sharp edges is called 

diffraction. Difference between interference and 

diffraction. 

Interference Diffraction 

(i) Two coherent sources (i) One source is 

are necessary 

Necessary 

(ii) All fringes are of (ii) CM has double 

same width 

width than all 

other fringes 

(iii) All bright fringes has (iii) as order of bright 

equal intensity 

fringes increases, 

intensity goes down 

(iv) For bright fringes (iv) For bright fringes, 

path difference = nλ Path difference 

λ 

= (2n – 1) . 

2 

(v) For dark fringes, path (v) For dark fringes, 

difference = (2n – path difference 

λ 

1) 

2 

= nλ 

Angular width of CM = 

2λ 

2× 

5000× 

10 

= 

−3 

a 0. 

5× 

10 

= 2 × 10 –3 Radian 

OR 

When white light incidents over prism then it divide 

it into component colours. This phenomenon is 

called diffraction and angle between end colours is 

called angular dispersion. Prism deviates violet 

colour maximum as its refractive index for violet 

colour is maximum. 


ω = 

µ v − µ 

µ −1 

y 


5318 −1. 

5140 

∴ ω = 


5170 −1 

∴ ω = .034 

29. white 

Types of spectrum 

Spectrum is of 2 types. 

(i) Emission 

(ii) Absorption. 

R 

−10

(i) Emission : - If radiations emitted from a substance 

is obtained ones screen after passing through the prism 

then spectrum is called emission spectrum. 

It is line emission spectrum for substances at atomic 

level in gaseous state. 

It is continuous spectrum for substances in liquid or 

solid state. 

(ii) Absorption spectrum : - The substance of which 

absorption spectrum is to be find out is placed in a 

transparent tube and white light is passed through it 

now substance absorb radiation corresponding to some 

particular wavelength and black lines are obtained for 

these absorbed radiations. These black lines are called 

absorption spectrum. 

OR 

Energy of emitted photons from H-atom = 10.2 eV 

Work – function of metallic surface 

12400 

= eV = 3.1 eV 

4000 

Now According to Einstein's Law of Photo-electric 

effect ' 

E = w + K.E.max. 

10.2 = 3.1 + K.E.max 

∴ K.E.max = 7.1 eV Ans 

30. Moving Coil Galvanometer – Principle : When a 

coil carrying current is placed in a magnetic field, it 

experiences a Torque τ = NiAB sin α. This Torque 

is proportional to the current passed through it. 

T1 

Phosphor 

Bronze 

Copper coil 

T2 

N 

Spring 

Torsion Head 

Mirror 

Frame 

S 

Core 

Working : When a current is passed through the 

coil, the two vertical limbs experience a force normal 

to it, equal in magnitude and parallel but opposite in 

sense. No forces act on the horizontal sides as they 

are parallel to the field .Since the field is radial, the 

forces acting on the vertical parts of coil remain 

always perpendicular to the plane of the coil in all its 

positions so that the perpendicular distance between 

the forces is always equal to breadth of the coil. Thus 

the coil is subjected to a torque whose magnitude is 

given by 

τ = NiAB … (i) 

Under the action of this deflecting torque, the coil 

rotates on its axis. Due to elastic forces, a restoring 

torque τr is produced in the suspension coil, which is 

perpendicular to the twist θ produced in the wire. 

τr = θ or τr = Cθ … (ii) 

where C is restoring torque per unit twist, and it is 

called the constant of twist for the suspension. As 

soon as restoring torque becomes equal to the 

deflecting torque, equilibrium is established. The coil 

does not rotate further. Let φ be the angle through 

which the coil rotates till it reaches the equilibrium 

position. 

τ = τr or NiAB = Cφ … (iii) 

C 

i = φ = K.φ 

NAB 

C 

K = is constant for the instrument, called 

NAB 

Galvanometer constant. 

∴ i ∝ φ . 

Thus deflection of coil is proportional to the current 

passing through it. The deflection can be measured 

by using a lamp and scale arrangement. 

OR 

Consider two infinitely long thin conductors 

carrying currents in opposite directions. 

Magnetic field B1 due to I1 at P on conductor CD is 

given by 


B1 = 

µ 0I1 

2πr 

F1B2 

B I1 

Q 

r 

D 

P 

B1F2 

A 

C 

The magnetic field B1 is perpendicular to plane of 

paper and directed inward. This field will produce a 

force/length F2 on conductor given CD by 

I2

µ 0I1I 

2 

F2 = B1I2 = 

[Q F = BI l, Here l = 1] 

2πr 

By Fleming's left hand rule direction of F2 is away 

from conductor AB. 

Similarly the current I2 will create a field B2 at Q 

directed inward which in turn will create force/length 

F1 

µ 0I1I 

2 

F1 = B2I1 = 

2πr 

By Fleming's left hand rule, the direction of F1 is 

away from the conductor CD. Hence the two 

conductors repel each other. 

Definition of Ampere If I1 = I2 = 1A, and r = 1m, 

then 

−7 

0 4π× 

10 

−7 

−1 

2 10 Nm 

µ 

F = = = × 

2π 

2π 

Thus one ampere is that current which on flowing 

through each of the two parallel uniform linear 

conductors placed in free space at a distance of one 

meter from each other produces between them a 

force of 2 × 10 –7 N per metre of their lengths. 

CHEMISTRY 

1. CH3–CH2–CH=CH2 + HCl → 

CH − CH 

2. (ii) > (iii) > (i) > (iv) 

3 

2 

− CH − CH 

3. By fehling solution, tollen reagent etc. 

4. Amino acid are compound containing amino group 

and carboxylic group. 

Structure of alanine is – 

+ 

H3N–CH–COO – 

CH3 

5. Movement of colloidal particles towards opposite 

terminal takes place which is called as 

electrophoresis. 

6. It is a stoichiometric defect in which an ion (cation) 

get displaced from its position and get arranged as an 

interstitial particle it is called Frenkel defect in this 

defect a vacancy or interstitial defect are formed 

simultaneously. Due to Frenkel defect. 

(i) Density remains same 

(ii) Electrical conductivity improved 

7. No of particle are different in sugar and NaCl. As 

dissociation in NaCl takes place. 

8. Methyl 3-bromo 2-oxo butanoate 

| 

Cl 

3 

9. (i) Sandmayer reaction. 

(ii) Kuchrous reaction 

CH ≡ CH ⎯→ CH2 = CH 

| 

OH 


—Cl 

10. (i) C H − N= 

C + 2HOH 

⎯⎯→ 

(ii) 

2 5 

Ethyl isocyanide 

Aniline 

H 

H 

NH 

+ Cl–C–CH3 

Acetyl chloride 

H 

+ 

Hydrolysis 

CH3 – CHO 

C2 

H5NH2 

+ HCOOH 

Ethyl amine 

H 

∆ 

Base 

O 

N–C–CH3 + HCl 

N-Phenylethanamide 

11. The given reaction will be as 

O 

H2O 

CH3CH2–C–H + CH3MgBr CH3CH2–CH–CH3 

(A) 

OH 

12. 

CH3–CH–CH–CH3 

Br2 CH3CH=CH–CH3 

Br Br 

(B) 

(C) 

Alc. KOH 

CH3–C≡C–CH3 

(D) 

O 

| | 

Structural formula of A : CH CH − C− 

H 

Structural formula of B : CH3CH=CH–CH3 

Structural formula of C : CH3–CH–CH–CH3 

Br Br 

Structural formula of D : CH3–C≡C–CH3 

(i) 

NH2 

Aniline 

+ CH3Cl + 3 KOH 

Chloroform 

NC 

Carbylamine 

3 

2 

∆ 

+ 3 KCl + 3 H2O 

–H2O 

H2SO4

OH 

(ii) + Br2 

Phenol 

H2O 

13. I st Method : Given K = 2.303 

2 . 303 × 0. 

3 

= 

K 

10 

t90 = × t1/2 

3 

∴ t1/2 = 

II method 

2. 

303 ⎛ a ⎞ 

K = ln ⎜ 0 ⎟ 

t ⎜ ⎟ 

⎝ a 0 − x ⎠ 

2.303 = 

t90 = 1 sec 

2. 

303 

t 90 

0 0 

14. P = Ax 

A PBx 

B 

100 

ln 

10 

Br 

OH 

Br 

Br 

2,4,6-tribromophenol 

2 . 303× 

0. 

3 

2. 

303 

= 0.3 sec. 

10 

= × 0.3 = 1 sec 

3 

P + {PºA = 450 mm 

PºB = 700 mm} 

600 = 450 xA + 700 xB 

= 450 (1 – xB) + 700 xB = 450 + 250 xB 

250 xB = 600 – 450 = 150 

3 2 

xB = & xA = 

5 5 

15. vapour pressure of solution pA = 0.8 pA 

W 

let mass of solute be Wg moles of solute = 

40 

114 

mole of octane n0 = = 1 

114 

W / 40 

xB = 

W / 40 + 1 

∆pA 

pA 

° – 0. 

8p 

A ° 

= 

= xB = 

p ° p ° 

A 

0.2 = 

A 

W / 40 

W / 40 + 1 

, W = 

W / 40 

W / 40 + 1 

0 . 2× 

40 

0. 

8 

= 10g 

16. (i) Aldol condensation : Two molecules of an 

aldehyde or a ketone having at least one 

α-hydrogen atom, condense in the presence of a 

dilute alkali to give β-hydroxy aldehyde or 

β-hydroxy ketone. 

O 

OH 

CH3–C + HCH2CHO 

Dil. NaOH 

CH3–C–CH2CHO 

H 

H 

Ethanal Ethanal Aldol 

(ii) Trans-Esterification : An ester on reaction with 

excess of alcohol in the presence of mineral acid 

forms a new ester. 

O 

CH3–C–OCH3 + CH3CH2OH 

Methyl Ethanol 

ethanoate 

O 


17 

(i) 

(ii) 

Benzene 

H 

CH3 – C = O 

Ethanal 

CH3COCl + Anhyd. AlCl3 

F.C.acylation 

CH3 – C = O + H2 

CH3 

Propanone 

H + 

CH3–C–OCH2CH3 + CH3OH 

Ethyl ethanoate Methanol 

CH3MgBr 

Dry ether 

Cu 

573 K 

COCH3 

Acetophenone 

H 

CH3 – C – OMgBr 

H + /H2O 

CH3 

H 

CH3 – C – OH 

18. (a) Cr +3 , µs = 15 Fe +3 > Cr +3 > V +3 

V +3 , µs = 8 

Fe +3 , µs = 35 

(b) CuSO4 (aq) 

CH3 

19. Conversions : 

(i) 1-bromopropane to 2-bromopropane 

CH3CH2CHBr alc. KOH CH3CH = CH2 

(ii) CH3–C–CH3 

(iii) 

OH 

O 

Phenol 

Na 

I2 

ONa 

O 

CH3–C–C.I3 

CO2 

4–7 atm. 

CH3CHCH3 

Br 

HBr 

(Peroxide) 

NaOH 

OH 

CHI3 

COONa

20. DNA has a hydrogen bonded double helical 

structure. The two strands are antiparallel. It can be 

considered a polymer of nucleotide (Base + sugar + 

phosphoric acid). In a nucleotide base and 

phosphate are linked to C1′ and C5 of sugar 

molecule respectively. Two nucleotide are linked 

by 3 ′ – 5′ phosphodiester bonds. Hydrogen bonds 

are formed between – 

Puric bases 

Adenine = Thyonine & Guanine ≡ cytosine 

5 ′ 

P 

S 

P 

3 ′ S 5 1 

4 3 2 

21. (a) CH2–CH 

Cl H 

(b) 

CH3 

CH2–C 

H 

OCOCH3 

(c) CF2–CF2 H 

Pyrimidal bases 

H bond 

A = T 

C ≡ G 

22. Antacids are drugs used to relieve acidity these can 

act in any of the following ways – 

(a) neutralize acid in stomach 

Ex – NaHCO3, milk of magnesia 

(b) stop acid production in stomach 

Ex – oneprazole, lasoprazole 

(c) antihistaminic 

Ex- Ranitidine (zantac) 

23. Absorption at the surface is called adsorption. In 

adsorption the conc. of adsorbate is greater at the 

surface than in the inner bulk. 

Adsorption of gases : 

Temperature : As temperature ↑ physisorption 

continuously ↓ where as with increase in 

temperature chemisorption first increases than 

decreases. 

Physisorption 

chemisorption 

T 

T 

Pressure : With increases in pressure adsorption of 

gases increases. 

S 

S 

P 

P 

3 ′ 

5 ′ 

24. (i) Schottky defect : It is a stoichiometric defect in 

which pair of cations and anions get displaced from 

their position in such a manner a pair of vacancies 

are created simultaneously 

In this defect 

(1) Density ↓ (2) Electrical conductiriy↑ 

(ii) Interstials : when particles may be an additional 

particle of same material a foreign particle get 

arranged in the spaces between regular arrangement 

of particles. 

(iii) F-centres : when crystal of NaCl is heated in the 

vapours of Na Cl – ions get displaced from its 

position and diffused into the Na vapour. And NaCl 

is formed which get deposited at the surface and in 

place of Cl – ion e – get arranged. This results in 

colour in the NaCl crystal. Other example LiCl in Li 

vapours become Pink, KCl in K vapours become 

violet. 

25. (i) Let a0 = 100 M, conc. after 10 min a0 – x 

= 100 – 20 

2. 

303 ⎛ a ⎞ 

k = log ⎜ 0 ⎟ 

t ⎜ ⎟ 

⎝ a 0 – x ⎠ 

2. 

303 100 –1 

= log = 0.02231 min 

10 80 

(ii) Let the time for 75% completion = t min 

x = 75 M 

2. 

303 ⎛ a ⎞ 

∴ t = log ⎜ 0 ⎟ 

k ⎜ ⎟ 

⎝ a 0 – x ⎠ 


= 

2. 

303 

0. 

02231 

100 

log = 62.15 min 

25 

26. (a) Cerium (Ce), 

(b) Due to lanthanoide contraction. 

(c) In basic medium, K2Cr2O7 converts into K2CrO4 

hence its colour changes while in acidic medium 

K2CrO4 converts into K2Cr2O7 

27 (a) 4FeCr2O4+16 NaOH+ 7O ⎯→ 

2 

( air) 

8 Na2CrO4+2Fe2O3 + 8H2O 

2Na2CrO4 + H2SO4 ⎯→ 

Na2Cr2O7 + Na2SO4 + H2O 

Na2Cr2O7 + 2KCl ⎯→ K2Cr2O7 +2NaCl 

(b) K2Cr2O7 + 7H2SO4 + 6KI ⎯→ 

Cr2(SO4)3 + 4K2SO4 + 3I2 + 7H2O 

2KMnO4+ 3H2SO4 + 5H2S ⎯→ 

2MnSO4 + K2SO4 + 8H2O + 5S

28. (a) Acetylene is first oxidized with 40% H2SO4 in the 

presence of HgSO4 

H – C ≡ C – H + H2O 

40%, 

H SO 

⎯ CH3 – CHO 

2 4 ⎯⎯⎯⎯→ 1% 

HgSO4 

Acetylene Acetaldehyde 

Acetaldehyde is finally oxidized to acid with air in the 

presence of manganous acetate catalyst 

Manganous 

CH3CHO + [ O] 

⎯⎯⎯⎯ 

⎯ →CH3COOH 

Acetaldehyde 

Acetate Acetic acid 

Ca( 

OH) 

2 

(b) (i) CH3COOH ⎯⎯⎯⎯→ 

(CH3COO)2 Ca 

∆ 

⎯⎯ 

Ca 

⎯⎯⎯→ 

(HCOO) 2 

Ca( 

OH) 

Calcium acetate 

CH 3CHO 

+ 2CaCO3 

Acetaldehyde 

2 

(ii) CH3COOH ⎯⎯⎯⎯→ 

(CH3COO)2Ca 

Acetic acid 

⎯⎯→ CH 3COCH3 

+ CaCO3 

Acetone 

∆ 

(c) When heated with I2 + NaCO3 Solution, acetone 

gives yellow crystals of iodoform CH3COCH3 + 

3NaOI → CH3I + CH3COONa 

Acetone Yellow ppt. 

(Iodoform) 

Acetic acid does not give iodoform test. 

(d) The carbonyl group in – COOH is inert and does 

not show nucleophilic addition reaction like carbonyl 

compounds. It is due to resonance stabilization of 

carboxylate ions. 

R – C = O 

R – C = O – 

O – 

Or 

(a) (i) Due to smaller + I-effect of one alkyl group 

in aldehydes as a compared to larger +I-effect of two 

alkyl groups, the magnitude of positive charge on the 

carbonyl carbon is more in aldehydes than in ketones. 

As a result nucleophilic addition reaction occur more 

readily in aldehyde than in ketones. 

(ii) The boiling points of aldehydes and ketones are 

lower than corresponding acids and alcohols due to 

absence of intermolecular hydrogen bonding . 

(iii) Aldehydes and ketones undergo a number of 

addition reactions as both possess the carbonyl 

functional group which reacts a number of 

nucleophiles such as HNC, NaHSO3, alcohols, 

ammonia derivatives and Grignard reagents. 

O 

(b) (i) Distinction between acetaldehyde and 

benzaldehyde: Acetaldehyde and benzaldehyde can be 

distinguish by Fehling solution. 

Acetaldehyde give red coloured precipitate with 

Fehling solution while benzaldehyde does not. 

2+ 

− 

4 

Fehling Solution 

CH 3CHO 

+ 2Cu 1442+ 

45 OH3⎯⎯→ 

CH COO Cu O H 2O 

3 + + 

red ppt. 

2 

(ii) Distinction between Propanone and Propanol : 

Propanone (CH3COCH3) and propanol 

(CH3CH2CH2OH) can be distinguish by iodoform 

test. Propanone when warmed with sodium 

hypoiodide (NaOI) i.e. I2 in NaOH, it gives yellow 

ppt of idoform while propanol does not respond to 

iodoform test. 

COCH 

+ 3NaOI 

⎯⎯→ 


CH 

3 3 

Pr opanone 

CH3 3 

Yellow ppt 

I ↓ + CH COONa + 2NaOH 

29. (a) Net cell reaction 

Mg (s) + 2Ag + (aq) → 2Ag(s) + Mg +2 (aq) 

According to nernst equation 

0. 

0591 

Ecell = E°cell – log 

n 

[ ] 

[ ] 2 

+ 2 

Mg 

+ 

Ag 

0. 

0591 0. 

2 

= 0.80 – (– 2.37) – log 

– 3 2 

2 ( 1× 

10 ) 

= 3.32 V 

when conc. of Mg +2 is decreased to 0.1, the new emf is 

0. 

0591 0. 

1 

Ecell = 3.17 – log 

– 3 2 

2 ( 1× 

10 ) 

= 3.34 V 

(b) (i) Mg > Αl > Zn > Fe > Cu 

As we move downwards in the electro chemical 

series tendency to displace increases. 

(ii) K > Mg > Cr > Hg > Ag 

30. (i) Cl2 > Br2 > F2 > I2 (due to exceptionally small size 

of F, F2 has lower bond energy than expected) 

(ii) HF < HCl < HBr < HI (according to bond length) 

(iii) NH3 > PH3 > AsH3 > SbH3 

(according to density of lonepair on central atom) 

(iv) H2O > H2Te > H2Se > H2S 

(according to strength of intermolecular bonding) 

(v) HOCl < HClO2

MATHEMATICS 

1. y = A cos (x + B) 

dy 

= – A sin (x + B) 

dx 

d 

2 

y 

2 

dx 

d 

2 

dx 

y 

2 

= – A cos (x + B) 

= – y ⇒ 

2 

d 

2 

dx 

Section A 

y 

2 

2 

+ y = 0 

2. ∫ cos x / 4 + sin x / 4 + 2 sin x / 4 cos x / 4 dx 

3. 

∫ (cos x / 4 + sin x / 4) 

dx = 4 sin (x/4) – 4 cos (x/4) + c 

→ 

a = iˆ 

– 2 ˆj 

+ 3kˆ 

; → 

b = iˆ 

– 3kˆ 

, 

⇒ 2 → 

a = 2 iˆ 

– 4 ˆj 

+ 6kˆ 

∴ → 

b × 2 → 

a = 

iˆ 

ˆj 

kˆ 

1 

2 

0 

– 4 

– 3 

= – 1 2iˆ 

– 12 ˆj 

– 4kˆ 

= – 4( 

3i 

ˆ + 3 ˆj 

+ kˆ 

) 

⇒ | → 

b × 2 → 

a | = | – 4( 

3i 

ˆ + 3 ˆj 

+ kˆ 

) | 

2 

2 

= 4 3 + 3 + 1 = 4 19 

4. → 

a = iˆ 

+ 2 ˆj 

– 3kˆ 

, → 

b = 3 iˆ 

– ˆj 

+ 2kˆ 

2 

⇒ → 

a + → 

b = 4 iˆ 

+ ˆj 

– kˆ 

, 

→ 

a – → 

b = 2 iˆ 

+ 3 ˆj 

– 5kˆ 

( → 

a + → 

b ) . ( → 

a – → 

b ) = ( 4 iˆ 

+ ˆj 

– kˆ 

).( 2 iˆ 

+ 3 ˆj 

– 5kˆ 

) 

= 4 × (– 2) + 1 × 3 – 1 × (– 5) 

= – 8 + 3 + 5 = 0 

→ 

⇒ a + → 

b is perpendicular to → 

a – → 

b . 

5. Let the position vectors of A, B, C, D be j ˆ 6 î – 7 , 

kˆ 16 î – 29 jˆ – 4 , kˆ 3 jˆ – 6 and kˆ 2 î + 5jˆ 

+ 10 

respectively. Then 

AB = OB – OA 

= ( kˆ 16 î – 29 jˆ – 4 ) – ( j ˆ 6 î – 7 ) 

= kˆ 10 î – 22 jˆ – 4 

6 

AC = ( kˆ 3 jˆ – 6 ) – ( j ˆ 6 î – 7 ) 

= – kˆ 6 î + 10 jˆ – 6 

AD = ( kˆ 2 î + 5jˆ 

+ 10 ) – ( j ˆ 6 î – 7 ) 

= kˆ – 4î 

+ 12 jˆ + 10 

Now ∆ = 


10 

– 6 

– 4 

– 22 

10 

12 

– 4 

– 6 

10 

Operate R1 → R1 + R2 + R3 

= 

0 

– 6 

– 4 

0 

10 

12 

0 

– 6 

10 

= 0 

⇒ The points A, B, C and D are coplanar. 

6. Put x = a sin θ 

= tan –1 ⎡ a sin θ ⎤ 

⎢ ⎥ 

⎣a 

cosθ 

⎦ 

= tan –1 (tan θ) = θ 

= sin –1 ⎛ x ⎞ 

⎜ ⎟ 

⎝ a ⎠ 

7. g (f(x)) = |5 f(x) – 2| 

⎡| 

5x 

– 2 |, 

= | 5 | x | – 2| = ⎢⎣ 

| – 5x 

– 2 |, 

8. ( B ) 

A × 

x ≥ 0 

x < 0 

3× 

4 ⋅ 4 2 . C2×3 

= (X)3×2 . C2×3 

= (Y)3×3 ∴ final order = 3 × 3 

9. x + 2 = – (2 × –3) ⇒ 3x = 1 ⇒ x = 1/3 

10. 2 – 20 = 2x 2 – 24 ⇒ 2x 2 = 6 ⇒ x = ± 3 

Section B 

11. Let A and B denote the two events respectively 

5 

P( A ) = 

5 + 2 

= 5 6 

, P(B) = 

7 6 + 5 

= 6 

11 

5 2 6 5 

P(A) = 1 – = , P( B ) = 1 – = 

7 7 

11 11 

P(At least one of A and B happens) 

= 1 – P(none of A and B happens) 

= 1 – P ( A ∩ B ) 

5 5 52 

= 1 – × = 

7 11 77

OR 

Let A → total of 8 in first throw 

B → total of 8 in 2nd throw 

Number of exhaustive cases when a pair of dice is 

thrown = 6 × 6 = 36 

Cases favourable to a total of 8 in each throw are 

(2, 6), (6, 2), (3, 5), (5, 3), (4, 4) 

Their number = 5 

P(A) = P(B) = 5/36 

5 5 25 

P(A and B) = P(A). P(B) = × = 

36 36 1296 

12. dx 

2x 

dy + 2 

1+ 

x 

2 

4x 

y = 2 

1+ 

x 

2x 

4x 

Here P = , Q = 2 

2 

1+ 

x 1+ 

x 

2x 

∫ dx 

+ 

2 

I.F. = e 1 x = e 

2 

ln( 

1+ 

x ) 

2 

= 1 + x 2 

Solution is : y (1 + x 2 2 

4x 

) = ∫ 2 

+ 

1 

y(1+x 2 4x ) = 

3 

3 

+ c 

x ⎡( 

2 + x) 

– 1⎤ 

13. ∫ e ⎢ ⎥ dx 

2 

⎣ ( 2 + x) 

⎦ 

∫ ⎥ ⎥ 

⎡ 

⎤ 

x 1 ⎛ – 1 ⎞ 

e ⎢ + ⎜ ⎟ 

⎜ ⎟ 

dx 

2 

⎢⎣ 

2 + x ⎝ ( 2 + x) 

⎠⎦ 

↓ ↓ 

f(x) f ′(x) 

1 

= e . c 

2 x 

x 

+ 

+ 

2x 

∫ 2 2 

1− 

x − ( x ) 

2 

dx 

OR 

x 

(1 + x 2 ) dx + c 

Let x 2 = t. Then, d(x 2 dt 

) = dt ⇒ 2x dx = dt ⇒ dx = 

2 x 

∴ I = ∫ 

dt 

2 

1− t − t 

= ∫ − { + −1} 

2 

dt 

t t 

= ∫ 

1 1 

− { + + − −1} 

4 4 

2 

dt 

t t 

⇒ Ι = ∫ 

⎪⎧ 

⎛ 1 ⎞ 5⎪⎫ 

− ⎨⎜ 

+ ⎟ − ⎬ 

⎪⎩ ⎝ 2 ⎠ 4⎪⎭ 

2 

dt 

= ∫ 

t 

dt 

2 

5 ⎛ 1 ⎞ 

− ⎜t 

+ ⎟ 

4 ⎝ 2 ⎠ 


= ∫ 

⎛ 

⎜ 

⎝ 

2 

5 ⎞ 

⎟ 

2 ⎟ 

⎠ 

⇒ I = sin –1 

dt 

⎛ 1 ⎞ 

− ⎜t 

+ ⎟ 

⎝ 2 ⎠ 

2 

⎛ t + 1/ 

2 ⎞ 

⎜ 

⎟ + C 

⎝ 5 / 2 ⎠ 

= sin –1 ⎛ 2t + 1⎞ 

⎜ 

⎟ + C 

⎝ 5 ⎠ 

= sin –1 ⎛ ⎞ 

⎜ 

2 + 1 

⎟ 

⎜ ⎟ 

⎝ 5 ⎠ 

2 

x 

+ C 

14. Divide by cos 2 x in N r & D r both 

∫ + dx 

x 

(tan x – 2)( 

2tan 

x 1) 

Let tan x = t 

sec 2 x dx = dt 

dt 

∫ ( t – 2)( 

2t 

+ 1) 

sec 2 

1 dt 

2 ∫ ( t – 2)( 

t + 1/ 

2) 

1 2 ⎡ 1 1 ⎤ 

. ∫ ⎢ – ⎥ dt 

2 5 ⎣ t – 2 t + 1/ 

2⎦ 

1 ⎡ 

⎢log 

5 ⎣ 

( t – 2) 

⎤ 

⎥ + c 

( t + 1/ 

2) 

⎦ 

15. Position vector of A is 

kˆ 3 î + jˆ + 2 ⇒ A (3, 1, 2) 

Position vector of B is 

kˆ î – 2 jˆ − 4 ⇒ B(1, –2, – 4) 

d.r.s of the line AB are 3, –1, 1 – (– 2), 2 – (– 4) 

i.e. 2, 3, 6 

∴ Eq. of the required plane through B and 

perpendicular line AB is 

2 (x – 1) + 3 (y + 2) + 6 (z + 4) = 0 

⇒ 2x + 3y + 6z + 28 = 0 

16. → 

a = kˆ î – 3jˆ 

+ , → 

b = kˆ î – jˆ + and → 

c = kˆ 2 î – jˆ – 

→ 

b × → 

c = 

î 

1 

2 

jˆ 

– 1 

– 1 

k ˆ 

1 

– 1 

∴ L.H.S = → 

a × ( → 

b × → 

c ) 

= kˆ 2 î + 3jˆ 

+

= 

î 

1 

2 

jˆ 

– 3 

3 

Next → 

a . → 

→ 

a . → 

k ˆ 

1 

1 

= kˆ – 6î 

+ jˆ + 9 

c = ( kˆ î – 3jˆ 

+ ).( kˆ 2 î – jˆ – ) 

= 2 + 3 – 1 = 4 

c = ( kˆ î – 3jˆ 

+ ). ( kˆ î – jˆ + ) 

= 1 + 3 + 1 = 5 

∴ R.H.S = ( → 

a . → 

c ) → 

b – ( → 

a . → 

b ) → 

c 

= 4( kˆ î – jˆ + ) – 5( kˆ 2 î – jˆ – ) 

= kˆ – 6î 

+ jˆ + = L.H.S. 

dy 

17. = 

dx 2 

3 

= 2 

3x 

– 2 

(slope of line) 

3 

⇒ = 

4 

41 

3 x – 2 ⇒ 9/16 = 2x – 2 ⇒ x = 

48 

y = 3/4 

eq n of tangent 

3 ⎛ 41 ⎞ 

(y – ) = 2 ⎜ x – ⎟⎠ ⇒ 48 x – 24 y = 23 

4 ⎝ 48 

18. Let y = 

y 

log y = log x 

2 

dy ⎛ 1 1 ⎞ 

dx 

⎜ – log x ⎟ 

⎝ y 2 ⎠ 

dy 

= 

dx 

y 

x ⇒ y = x y/2 

y 

x( 

2 – ylog 

x) 

2 

Let y = tan –1 

⎡ 

⎢ 

⎢ 

⎣ 

put x 2 = cos 2θ 

y = tan –1 

⎪⎧ 

⎨ 

⎪⎩ 

y 

= 

2x 

1+ 

x 

1+ 

x 

2 

2 

OR 

− 

+ 

1+ 

cos 2θ 

− 

1+ 

cos 2θ 

+ 

⎡cosθ 

− sin θ⎤ 

y = tan –1 ⎢ ⎥ 

⎣cosθ 

+ sin θ⎦ 

y = tan –1 [tan (π/4 –θ)] 

π 1 –1 2 

y = – cos x 

4 2 

1− 

x 

1− 

x 

2 

2 

1− 

cos 2θ 

⎪⎫ 

⎬ 

1− 

cos 2θ 

⎪⎭ 

⎤ 

⎥ , z = cos 

⎥ 

⎦ 

–1 x 2 

dz 

= 

dx 

−1 

1− 

x 

4 

.2x 

dy −1 

⎛ 

= ⎜ 

dx 2 ⎜ 

⎝ 


dy 

= 

dx 

dy 

∴ = 

dz 

x 

1− x 

−1 

1− 

x 

4 

4 

⎞ 

. 2x 

⎟ 

⎟ 

⎠ 

dy / dx x / 1− 

x 

= 

dz / dx − 2x 

/ 1− 

x 

19 Taking log 

y log x = x – y 

x dy ( 1+ 

y = ⇒ = 

1+ 

log x dx 

dy = 2 

dx 

( 1 

log x 

+ 

log x) 

20. Q f(π/4) = f(π/4 + ) 

= 

= 

h 0 

lim → 

4 

( 1 

4 

− 1 

= 

2 

log x). 

1– 

+ 

tan[ π/ 

4 – ( π/ 

4 + h)] 

cot 2( 

π/ 

4 + h) 

– tan 

h 

h 0 

lim → – tan 2h 

= 

h 0 

lim → 

x 

2 

log x) 

21. Let y = f(x) 

y = 1 + αx 

y – 1 

x = ∴ f 

α 

–1 x – 1 

(x) = 

α 

Now f(x) = f –1 (x) 

x – 1 

1 + αx = 

α 

Compose 1 = –1/α, α = 1/α 

α = – 1 , α = ± 1 

22. L.H.S = 

= x 2 

1 

5 

10 

= x 2 

⎡ 1 

⎢ 

⎢ 5 

⎢ 

⎣ 10 

x 

5x 

10x 

1 

4 

8 

1 

4 

8 

α = – 1 

x x 

4x 

x 

8x 

x + 4 

x 

x + y.x 

x + 4 

x 

x 

x 

+ 

1 

5 

10 

1 

4 

8 

+ 

1 

4y 

8y 

( 1/ 

x) 

⎛ tan h ⎞ 

⎜ ⎟h 

⎝ h ⎠ 

= 1/2 

⎛ tan 2h 

⎞ 

⎜ ⎟2h 

⎝ 2h 

⎠ 

y 

4y 

8y 

1 

4x 

8x 

x x 

x 

x + 4 

x 

x 

4 + x 

4x 

8x 

0 ⎤ 

⎥ 

0 ⎥ + 0 

4 ⎥ 

⎦

= x 2 

⎡ 

⎢ 

⎢x 

⎢ 

⎣ 

0 

1 

2 

1 

3 

7 

1 ⎤ 

⎥ 

1 + 0 ⎥ + 0 = x 

1 ⎥ 

⎦ 

3 .(7 – 6) = x 3 

OR 

1 

1 

1 

x 

y 

z 

yz 

zx 

xy 

1 

= 

xyz 

x 

y 

z 

2 

x 

2 

y 

2 

z 

xyz 

xyz 

xyz 

xyz 

= 

xyz 

x 

y 

z 

x 

y 

z 

⇒ 

1 

1 

1 

x 

y 

z 

y 

y 

z 

2 

2 

2 

C1 ↔ C3 

R1 → R1 – R2, R2 → R2 – R3 

= 

0 

0 

1 

x − y 

y − z 

z 

x 

= (x – y) (y – z) 

y 

2 

2 

− y 

− z 

z 

0 

0 

1 

2 

2 

2 

1 

1 

z 

x + y 

y + z 

1 x + y 

= (x – y) (y – z) 

1 y + z 

= (x – y) (y – z) [(y + z) – (x + y)] 

= (x – y) (y – z) (z – x) 

z 

Section C 

23. The ball transferred from the first bag to the second 

bag can either be white or black. 

Case I : When white ball is transferred from the I st 

bag to the second. 

Probability of getting white ball from I st 5 

bag = 

5 + 4 

5 

= . 

9 

Now, second bag contains 8 white and 9 black balls. 

The probability of drawing a white ball from the 

8 

second bag = 

8 + 9 

= 8 

. 

17 

∴ The probability of both these events taking place 

5 18 40 

together is = × = 

9 17 153 

Case II : When a black ball is transferred from the I st 

bag to the second probability of getting black ball 

from I st 4 

bag = 

5 + 4 

= 4 

. 

9 

2 

2 

2 

2 

1 

1 

1 

The probability of drawing a white ball from the 

7 7 

second bag is = 

7 + 10 17 

Q The probability of both these events taking place 

4 7 28 

together is = × = 

9 17 153 

Hence, the required probability 

40 28 68 4 

= + = = . 

153 153 153 9 


24. 

required are 

(0, 0) (1, 0) 

1/ 

2 

A(1/2, 3 / 2 ) 

(2, 0) 

⎡ 

2 

2 ⎤ 

A = 2⎢ 

+ ⎥ 

⎣∫ 

1– 

( x – 1) 

dx ∫ 1– 

x dx 

0 

1/ 

2 ⎦ 

⎡⎛ 

1 

A = 2 ⎢⎜ 

( x – 1) 

⎢⎣ ⎝ 2 

⎛ 1 

+ ⎜ x 

⎝ 2 

25. I st Part 

π 

I = ∫ 

0 

e 

1– 

x 

e 

cosx 

By property 

π 

I = ∫ 

0 

e 

e 

– cosx 

2 

cosx 

+ e 

– cosx 

+ e 

⎡ 

By property ⎢ 

⎢ 

⎣ 

+ 

1– 

( x – 1) 

– cosx 

1 

sin 

2 

cosx 

π 

∫ 

0 

eq n (1) + (2) 

π cosx 

e + e 

2 I = ∫ cosx 

e + e 

0 

2 nd Part 

π/ 

4 

– 1 

2 

⎞ 

x⎟ 

⎠ 

1 

1 

+ 

1/ 

2 

1 

sin 

2 

⎤ 

⎥ 

⎥⎦ 

– 1 

1/ 

2 

⎞ 

( x – 1) 

⎟ 

⎠ 

2π 3 

= – 

3 2 

.... (1) 

.... (2) 

π ⎤ 

f ( x) 

dx = π ⎥ ∫ f ( – x) 

dx 

⎥ 

0 ⎦ 

– cosx 

– cosx 

dx ⇒ I = π/2 

π/ 

4 

dx 

I = ∫ dx + π/ 

4 

2 – cos2x 

∫ 

– π/ 

4 

– π/ 

4 

dx 

2 – cos2x 

0

I1 I2 

Since I1 is an odd function therefore I1 = 0 

π/ 

4 

I = π/ 

4 ∫ 

– π/ 

4 

π/ 

4 

= π/ 

4 ∫ 

2 

1 

– π/ 

4 

π/ 

4 

dx 

⎛ 

⎜ 

1– 

tan 

2 – 

⎜ 

⎝1 

+ tan 

+ ( 

sec 

2 

x 

2 

2 

x ⎞ 

⎟ 

x ⎟ 

⎠ 

2 

3 tan x) 

sec x 

= 2 π/ 

4 ∫ 

[Q f(x) is even function] 

2 

2 

1 + ( 3 tan x) 

0 

2 

Let 3 tan x = t ⇒ sec 2 dt 

x dx = 

3 

dt / 3 

= ∫ 2 2 1+ 

t 

π 3 

0 

= ( ) 3 – 1 

/ 2 3 tan t 

π = 

26. The given lines are 

→ 

r = ( kˆ î – 2jˆ 

+ 3 ) + t ( kˆ – î + jˆ – 2 ) 

and → 

r = ( kˆ î – jˆ – ) + s ( kˆ î + 2jˆ 

– 2 ) 

Here → 

1 

→ 

2 – 

0 

2 

π 

a = kˆ î – 2jˆ 

+ 3 ; → 

b 1 = kˆ – î + jˆ – 2 

a = kˆ î jˆ – ; 

→ 

b = kˆ î + 2 jˆ – 2 

b × → 

b = 

∴ → 

1 

2 

î 

– 1 

1 

jˆ 

1 

2 

k ˆ 

– 2 

– 2 

The S.D. between the gives lines 

= 

= 

→ 

→ 

( b1× 

b2 

).( a1 

– a 2 ) 

| b × b | 

2 

2 

1 

4 + 12 

+ 4 

2 

→ 

2 

2 

+ 3 

→ 

= 

8 

= 

29 

2 

6 

3 

= kˆ 2 î – 4 jˆ – 3 

k | ˆ 

k) | 2î 

– 4jˆ 

– 3 

ˆ k).( jˆ – 4 ˆ ( 2î 

– 4 jˆ – 3 

27. Part I st 

Let x kg and y kg. of fertilizer A and B be mixed by 

the farmer. Then LPP is 

Minimize : C = 5x + 8y 

subject to the constraints 

10 5 

x + y ≥ 7 ⇔ 2x + y ≥ 140 

100 100 

6 10 

x + y ≥ 7 ⇔ 3x + 5y ≥ 350 

100 100 

x ≥ 0, y ≥ 0 

y 

(1) 

(0, 140)B 

D 


O 

(50, 40) 

P 

A C 

⎡350 

⎤ 

⎢ , 0⎥ 

⎣ 3 ⎦ 

(2) 

Draw the lines 

2x + y = 140 ..... (1) 

3x + 5y = 350 ..... (2) 

The shaded unbounded region is the feasible region. 

These line meet at P(50, 40) 

Now value of c = 5x + 8y 

⎡350 

⎤ 1750 1 

at C ⎢ , 0⎥ 

, is = 583 

⎣ 3 ⎦ 3 3 

at P(50, 40) is 570 

at B(0, 140) is 1120 

∴ The cost is minimum at P(50, 40) 

This occurs when 50 kg of type & fertilizer and 40 

kg of type B fertilizer are mixed to the meet the 

requirement. 

Part 2 nd 

Let x and y represent the number of tables and chairs 

respectively that the dealer sells and P be the profit 

of the dealer. Then the required LPP is . 

Maximize: P = 50x + 15y 

Subject to the constraints 

500x + 200 y ≤ 10,000 ⇔ 5x + 2y ≤ 100 

x + y = 60 ....(1) 

5x + 2y = 100 .... (2) 

y 

D 

(0, 50) 

B(0, 60) 

A(60, 0) 

x 

O C (20, 0) 

(1) 

(2) 

A shaded region OCD is the feasible region. 

Now P = 50x + 15y 

At O, value of P is O 

At C, value of P is 50 × 20 = 1,000 

At D, value of P is 15 × 50 = 750 

∴ Profit is maximum at C (20, 0) 

x

28. 

x 

x 

y 

∴ x → side of square 

y → rad. of circle 

∴ 4x + 2πy = 36 ..... (1) 

Now area A = x 2 + πy 2 

A = x 2 ⎛18 

– 2x 

⎞ 

+ π ⎜ ⎟ [From (1)] 

⎝ π ⎠ 

dA 4 

= 2x – (18 – 2x) 

dx π 

2 

d A 

= 2 + 8/π > 0 ⇒ min 2 

dx 

m area 

Form min m dA 

area = 0 

dx 

36 

⇒ x = 

π + 4 

18 

From (1) , y = 

π + 4 

144 

Length of one piece = 4x = 

π + 4 

36π 

Length of other piece = 2πy = 

π + 4 

1 

29. Let = u ; 

x 

2 

1 1 

= v; = w 

y z 

Given equation can be written in form 

⎡ 2 3 10 ⎤ ⎡ u ⎤ ⎡4⎤ 

⎢ 

⎥ 

⎢ 

4 – 6 5 

⎢ ⎥ 

⎥ ⎢ 

v 

⎥ 

= 

⎢ ⎥ 

⎢ 

1 

⎥ 

⎢⎣ 

6 9 – 20 ⎥⎦ 

⎢⎣ 

w⎥⎦ 

⎢⎣ 

2⎥⎦ 

A ⋅ X = B 

X = A –1 ⋅ B 

Adj( 

A) 

X = . B, |A| = 1200 

| A | 

X = 

⎡ 75 

⎢ 

⎢ 

110 

⎢⎣ 

72 

150 

– 100 

0 

1200 

75 

30 

– 24 

⎤ ⎡4⎤ 

⎥ 

. 

⎢ 

1 

⎥ 

⎥ ⎢ ⎥ 

⎥⎦ 

⎢⎣ 

2⎥⎦ 

⎡600⎤ 

⎡ u ⎤ ⎢ 

X = 

⎢ ⎥ 400 

⎥ ⎡1/ 

2⎤ 

⎢ 

v 

⎥ 

= ⎢ ⎥ = 

⎢ ⎥ 

⎢240⎥ 

⎢ 

1/ 

3 

⎥ 

⎢⎣ 

w⎥ 

⎣ ⎦ 

⎦ 

⎢ ⎥ 

1200 

⎣1/ 

5⎦ 

u = 1/2 ⇒ x = 2 

v = 1/3 ⇒ y = 3 

w = 1/5 ⇒ z = 5 

At a Glance 

Some Important Practical Units 

1. Par sec : It is the largest practical unit of 

distance. 

1 par sec = 3.26 light year 

2. X-ray unit : It is the unit of length. 

1 X-ray unit = 10 –13 m 

3. Slug : It is the unit of mass. 

1 slug = 14.59 kg 

4. Chandra Shekhar limit : It is the largest 

practical unit of mass. 

1 Chandra Shekhar limit = 1.4 × Solar mass 

5. Shake : It is the unit of time. 

1 Shake = 10 –6 second 

6. Barn : It is the unit of area. 

1 barn = 10 –28 m 2 

7. Cusec : It is the unit of water flow. 

1 cusec = 1 cubic foot per second flow 

8. Match No. : This unit is used to express velocity 

of supersonic jets. 

1 match no. = velocity of sound 

= 332 m/sec. 

9. Knot : This unit is used to express velocity of 

ships in water. 

1 knot = 1.852 km/hour 

10. Rutherford : It is the unit of radioactivity. 

1 rutherford (rd) = 1 × 10 6 disintegrations/sec 

11. Dalton : It is the unit of mass. 

1 12 

1 dalton = mass of C = 931 MeV 

12 

= 1 a.m.u. 

12. Curie : It is the unit of radioactivity. 

1 curie = 3.7 × 10 10 disintegration / sec 


XtraEdge Test Series 

ANSWER KEY 

PHYSICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans C A A A A C C D A B 

Ques 11 12 13 14 15 16 17 18 19 

Ans B A D B B A B A D 

Column 

Matching 

20 A → P,R,S B → P,R,S C → P,Q D → P,Q 

21 A → R,S B → P C → R D → S 

22 A → P,Q,R,S B → P,R C → Q,R D → R 

CHEMISTRY 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans B A A C B B C A A B 

Ques 11 12 13 14 15 16 17 18 19 

Ans C A D B D A C C C 

Column 

Matching 

20 A → P,S B → S C → Q D →RP 

21 A → S B → P C → Q D → R 

22 A → P,Q B → P C → P,Q,R D → S 

MATHEMATICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans B D C B B C A D C D 

Ques 11 12 13 14 15 16 17 18 19 

Ans C D A A C B B C A 

Column 

Matching 

20 A → S B → Q C → P D → R 

21 A → S B → R C → P D → P 

22 A → R B → S C → Q D → P 

PHYSICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans C D A D A B A A D C 

Ques 11 12 13 14 15 16 17 18 19 

Ans A C B C B C C D B 

Column 

Matching 

IIT- JEE 2012 (February issue) 

IIT- JEE 2013 (February issue) 

20 A → R B → P,S C → P,Q D → Q 

21 A → Q B → P,R,S C → P,R D → S 

22 A → P,Q B → Q C → R,S D → S 

CHEMISTRY 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans A D C D A C B A A C 

Ques 11 12 13 14 15 16 17 18 19 

Ans A C C A A A B C A 

Column 

Matching 

20 

21 

22 

A → Q,R 

A → R 

A → P,S 

B → P,Q,R,S 

B → S 

B → P 

C → P,S 

C → Q 

C → Q 

D → P,Q,R 

D → P 

D → P,R,S 

MATHEMATICS 

Ques 1 2 3 4 5 6 7 8 9 10 

Ans A B B B A B A B D D 

Ques 11 12 13 14 15 16 17 18 19 

Ans C D D B C A A B D 

Column 

Matching 

20 

21 

22 

A → Q 

A → R 

A → R 

B → S 

B → P 

B → S 

C → R 

C → S 

C → P 

D → P 

D → Q 

D → Q 


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Every month get the XtraEdge Advantage at your door step. 

✓ Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE. 

✓ Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice. 

✓ Take advantage of experts' articles on concepts development and problem solving skills 

✓ Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda. 

✓ Confidence building exercises with Self Tests and success stories of IITians 

✓ Elevate you to the international arena with international Olympiad/ Contests problems and Challenging Questions. 

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