8 Questions<strong>Solution</strong>Set # 12Physics Challenging ProblemsPublished in this Issue1.[C] Magnetic field due to infinite current carryingµ Jsheet is given by B = 0, where J is linear current2density.II IIIµ 0 J2µ 0 J2µ 0 J2µ 0 J2(a) (b)Fig. (a) and (b) represent the direction of magneticfield due to current carrying sheets. For x < a,µ 0Jµ 0J(2J)µ 0 (3J) µ 0 (4J)Bresul tant = − − +2 2 2 2For a < x < 2a,µ 0Jµ 0 (2J) µ 0 (3J) µ 0 (4J)tant = − − + = −2 2 2 2For 2a < x < 3a,µ 0Jµ 0 (2J) µ 0 (3J) µ 0 (4J)Bresul tant = + − − = 02 2 2 2So, the required curve isBBresul µ 0Oa 2a 3a 4a 5aXJ(C) This situation is similar to part (i)(D) In a uniform electric field, path can be only3. A → Q B → R C → P D → Q(A) At t = 1s, flux is increasing in the inwarddirection, hence induced e.m.f. will be inanticlockwise direction.(B) At t = 5s, there is no change in flux, so inducede.m.f. is zero(C) At t = 9s, flux is increasing in upward directionhence induced e.m.f. will be in clockwise direction.(D) At t = 15s, flux is decreasing in upwarddirection, so induced e.m.f. will be inanticlockwise direction.4.[A, B, D]Rate of work done by external agent isde/dt = BIL.dx/dt = BILv and thermal powerdissipated in resistor = eI = (BvL) I clearly both areequal, hence (A).If applied external force is doubled, the rod willexperience a net force and hence acceleration. As aresult velocity increase, hence (B).Since, I = e/ROn doubling R, current and hence required powerbecome half.Since, P = BILvHence (D)2. A → P,Q,S B → P,Q,R,SC → P,Q,R,S D → Q(A) Velocity of the particle may be constant, if forcesof electric and magnetic fields balance eachother. Then, path of particle will be straight line.Also, path of particle may be helical if magneticand electric fields are in same direction. But pathof particle cannot be circular. Path can becircular if only magnetic field is present, or ifsome other forces is present which can cancel theeffect of electric field.(B) Here, all the possibilities are possible dependingupon the combinations of the three fields.5.[A]→∧^jxzn 1 sini = n 2 sinr→1.5(µ 1×j) = 2( µ 2×j)∧∧∧1.5(a i + b j) × j = 2[(c i + d j) × j]∧=∧1 .5a k 2c ka 20 4= =c 1.5 3∧∧∧∧XtraEdge for IIT-JEE 18 APRIL 2010
6.[A]I 2f–f sin60º f cos60º +fI 1xu = -f cos60ºf = +f1 1 1= −f v − f cos 60º1 1 2= +f v f1 2 1− =f f vv = -ff = cos 60ºxf= xcos 60ºx = 2f∴ final image will formed at optical centre of firstlens.7.[C] C v = (3 + 2T)RdQ = dU + PdVadiabatic process dQ = 00 = Rn (3 + 2T)dT + PdVnRT0 = Rn(3 + 2T)dT + dVVdV ⎛ 3 + 2T ⎞∫−= ∫⎜⎟dTV ⎝ T ⎠– log V = 3 logT + 2T + C– logV – logT 3 = 2T + Clog VT 3 = 2T + CVT 3 = e 2TVT 3 e -2T = C8.[A] P = P 0 − αVPV = RTRT2= P 0 − αVV3P0 V αVT = −R RdT = 0dV2P03αV− =R R0PV = 03 α2Now put V in T.• Saturn’s rings are made up of particles of ice, dustand rock. Some particles are as small as grains ofsand while others are much larger thanskyscrapers.• Jupiter is larger than 1,000 Earths.• The Great Red Spot on Jupiter is a hurricane-likestorm system that was first detected in the early1600’s.• Comet Hale-Bopp is putting out approximately250 tons of gas and dust per second. This isabout 50 times more than most comets produce.• The Sun looks 1600 times fainter from Pluto thanit does from the Earth.• There is a supermassive black hole right in themiddle of the Milky Way galaxy that is 4 milliontimes the mass of the Sun.• Halley’s Comet appears about every 76 years.• The orbits of most asteroids lie partially betweenthe orbits of Mars and Jupiter.• Asteroids and comets are believed to be ancientremnants of the formation of our Solar System(More than 4 billion years ago!).• Comets are bodies of ice, rock and organiccompounds that can be several miles in diameter.• The most dangerous asteroids, those capable ofcausing major regional or global disasters, usuallyimpact the Earth only once every 100,000 yearson average.• Some large asteroids even have their own moon.• Near-Earth asteriods have orbits that cross theEarth’s orbit. These could potentially impact theEarth.• There are over 20 million observable meteorsper day.• Only one or two meteorites per day reach thesurface of Earth.• The largest found meteorite was found in Hoba,Namibia. It weighed 60 tons.XtraEdge for IIT-JEE 19 APRIL 2010
- Page 3 and 4: Volume - 5 Issue - 10April, 2010 (M
- Page 5 and 6: Volume-5 Issue-10April, 2010 (Month
- Page 7 and 8: Hospital in Parel and ApolloHospita
- Page 9: KNOW IIT-JEEBy Previous Exam Questi
- Page 12 and 13: CHEMISTRY6. From the following data
- Page 14 and 15: (ii)OHH + CH 3OH⊕PorClPClClOCH 3O
- Page 16 and 17: Physics Challenging ProblemsSet #12
- Page 18 and 19: 8 QuestionsSolutionSet # 11Physics
- Page 22 and 23: PHYSICSSStudents'ForumExpert’s So
- Page 24 and 25: If the body loses this heat in time
- Page 26 and 27: Perfect gas equation :From the kine
- Page 28 and 29: The internal energy of n molecules
- Page 30 and 31: * The binding energy per nucleon is
- Page 32 and 33: KEY CONCEPTOrganicChemistryFundamen
- Page 34 and 35: KEY CONCEPTPhysicalChemistryFundame
- Page 36 and 37: UNDERSTANDINGOrganic Chemistry1. An
- Page 38 and 39: The starting compound (A) reacts wi
- Page 40 and 41: `tà{xÅtà|vtÄ V{tÄÄxÇzxá12Se
- Page 43 and 44: MATHEMATICAL CHALLENGESSOLUTION FOR
- Page 45 and 46: A 2 ⎛ 3s − 2s ⎞ s≤ s ⎜
- Page 47 and 48: 5. Let z 1 , z 2 and z 3 are unimod
- Page 49 and 50: If I n =∫x n . e ax dx, then I n
- Page 51 and 52: The common difference of an A.P. is
- Page 53 and 54: XtraEdge for IIT-JEE 51APRIL 2010
- Page 55 and 56: 4. Consider the following reactions
- Page 57 and 58: 15. What would be the structure of
- Page 59 and 60: 14. If line lies inside the region
- Page 61: 6. A BTwo containers A & B contain
- Page 64 and 65: MOCK TEST FOR IIT-JEEPAPER - IITime
- Page 66 and 67: SECTION - IVInteger answer typeThis
- Page 68 and 69: X Y Z W0123456789012345678912. If z
- Page 70 and 71:
(C) Pressure at A > pressureat B(D)
- Page 72 and 73:
7. Experimental verification of New
- Page 74 and 75:
28. Assertion : A rocket moves forw
- Page 76 and 77:
60. Point out the incorrect stateme
- Page 78 and 79:
MOCK TEST - BIT-SATTime : 3 Hours T
- Page 80 and 81:
22. The displacement of interfering
- Page 82:
40. A body of mass 2 kg is being dr
- Page 85 and 86:
10. The foci of a hyperbola coincid
- Page 87 and 88:
LOGICAL REASONING1. Fill in the bla
- Page 89 and 90:
SOLUTION FOR MOCK TESTIIT-JEE (PAPE
- Page 91 and 92:
or 0.1 v α + 1 = 2 + 2 × 10 -2 ×
- Page 93 and 94:
12.[A, B, C, D]13.[B]2x t − 5t +
- Page 95 and 96:
R l=sin 45º sin 30ºR sin 45º = =
- Page 97 and 98:
1.[C]2.[B]3.[B]SOLUTION FOR MOCK TE
- Page 99 and 100:
5.[A,B,D] We have adj A = |A| A -1a
- Page 101 and 102:
(1, 1)7λ 5.5× 10t min = =P4n 4×1
- Page 103 and 104:
K TFor a cylinder = 2 in case ofK R
- Page 105 and 106:
∴17.[C]Potential of shell A is,1
- Page 107 and 108:
39.[B] Solution is decinormal, that
- Page 109 and 110:
71.[B]which is a H.P.n th term of c
- Page 111 and 112:
85.[C] Letters of word 'STATISTICS'
- Page 113 and 114:
13. [B] Point P lies on the arms CD
- Page 115 and 116:
33. [B] For quarter revolution∆ V
- Page 117 and 118:
16.[A]NH 2⎯ 0−5°C+ NaNO 2 + 2H
- Page 119 and 120:
11.[B]→ →∴ α & β are two mu
- Page 121 and 122:
1/ 21 -2∫2 x 2 dx⇒⇒01 2 - 3 2
- Page 124 and 125:
8.[A]9.[A]10.[B]To take off :irrele
- Page 126:
CAREER POINTIn associationwithLaunc