May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
40. A block is moving on an inclined plane makingan angle 45º with the horizontal and thecoefficient of friction is µ. The force required tojust push it up the inclined plane is 3 times theforce required to just prevent it from slidingdown. If we define N = 10 µ, then N is.Ans. [5]Sol.45ºN=mgcosθmgsinθ + µmgcosθ = 3(mgsinθ – µmgcosθ)1 µ 3 3µ⇒ + = −2 2 2 2⇒4µµ = 212=22⎛ 1 ⎞N = 10µ = 10 ⎜ ⎟ = 5⎝ 2 ⎠41. A boy is pushing a ring of mass 2 kg and radius0.5 m with a stick as shown in the figure. Thestick applies a force of 2 N on the ring and rolls itwithout slipping with an acceleration of 0.3 m/s 2 .The coefficient of friction between the groundand the ring is large enough that rolling alwaysoccurs and the coefficient of friction between thestick and the ring is (P/10). The value of P is.StickAns. [4]Sol.mgGroundf 22NP×2 Pf 1 = µ×2 = =10 52 – f 2 = Ma cm …….(1)f 2 = 2 – 2 × 0.3 = 1 .4 N(f 2 – f 1 ) R = I cm αa(f 2 – f 1 ) R = MR2 cm×Rf 2 – f 1 = Ma cmf 1 = f 2 – ma cm = 1.4 – 2 × 0.3 = 0.8 NP0 .8 = ⇒ P = 45Note : It has been assumed that the stick applieshorizontal force of 2N (only normal reaction)42. Four point charge, each of +q, are rigidly fixed atthe four corners of a square planar soap film ofside 'a'. The surface tension of the soap film is γ.The system of charges and planar film are inequilibrium, andAns. [3]Sol.⎡2q ⎤a = k ⎢ ⎥⎦⎣ γF AC1/NF AB, where 'k' is a constant. Then N is.F +qADA+qDF AC =BC+q +qq28πεa0F AD = F AB =F R =⇒ a 3 =22q24πεa0q ⎛ 1 ⎞⎜2cos45º+ ⎟24πε0a⎝ 2 ⎠= r (2) (BD) = 2 r( 2 a)2⎛q ⎜⎝8⎛2q ⎞a = k⎜⎟r⎝ ⎠1/3⎛ 1 ⎞⎜ 2 + ⎟where k = ⎜ 2 ⎟⎜ 8 2π⎟⎝ ⎠21 ⎞2 + ⎟2 ⎠2πεr1/30⇒ N = 3XtraEdge for IIT-JEE 72 MAY 2011
43. Four solid spheres each of diameter 5 cm andmass 0.5 kg are placed with their centers at thecorners of a square of side 4 cm. The momentumof inertia of the system about the diagonal of thesquare is N × 10 –4 kg-m 2 , then N is.Ans. [9]Sol.mmmr =lm5 cm = × 1025 –2 m2m = 21 kgl = 4 × 10 –2 mUsing parallel axis theorem⎡ 2 1 5 −4 ⎤I total = ⎢4 × × × × 10 ⎥ +⎣ 5 2 4 ⎦⎡ 1 −4 ⎤⎢2× × 8×10 ⎥⎣ 2 ⎦⇒ 10 –4 + 8 × 10 –4 ⇒ 9 × 10 –4 kg m 244. The activity of a freshly prepared radioactivesample is 10 10 disintegrations per second, whosemean life is 10 9 s. The mass of an atom of thisradioisotope is 10 –25 kg. The mass (in mg) of theradioactive sample is.Ans. [1]Sol. A = λN10 10 1=9 × N10N = 10 19mass of sample = 10 19 × 10 –25 × 1 × 10 6 = 1 mg45. A long circular tube of length 10 m and radius 0.3carries a current I along its curved surface asshown. A wire-loop of resistance 0.005 ohm andof radius 0.1 m is placed inside the tube with itsaxis coinciding with the axis of the tube. Thecurrent varies as I = I 0 cos(300 t) where I 0 isconstant. If the magnetic moment of the loop is Nµ 0 I 0 sin(300 t), then 'N' is.Ans. [6]Sol. B =µ 0 0I cos300 t10µ Iφ = 0 0× 3.14 × 0.01 cos 300 t10φ = 3.14 × µ 0 I 0 cos 300 t × 10 –3dφe = − = 3.14 × 300 µ 0 I 0 sin 300t × 10 –3dtI−3e 3.14 × 300 µ I sin 300t 10i = =0 0 ×R 0. 005i = 3.14 × 60 µ 0 I 0 sin 300 t0.01Magnetic moment = 3.14 × 60 × 3.14 × × 100µ 0 I 0 sin 300 t= 5.9 µ 0 I 0 sin 300 t= 6 µ 0 I 0 sin 300 t46. Steel wire of length 'L' at 40ºC is suspended fromthe ceiling and then a mass 'm' is hung from itsfree end. The wire is cooled down from 40ºC to30ºC to region its original length 'L'. Thecoefficient of linear thermal expansion of the steelis 10 –5 /ºC, Young's modulus of steel is 10 11 N/m 2and radius of the wire is 1 mm. Assume that L >>diameter of the wire. Then the value of 'm' in kgis nearly.Ans. [3]Sol.mmgL∆L = = Lα(∆θ)AYAYα(∆θ)⇒ m ==gm = 3.14 kg ⇒ 3 kgπ × 10−611× 10 × 1010−5× 10XtraEdge for IIT-JEE 73 MAY 2011
- Page 24 and 25: the indivual forces. It's often hel
- Page 26 and 27: Hence, E x = -∴∂V∂ xE = - ayi
- Page 28 and 29: For (x - a) 2 + (y - b) 2 = r 2 , t
- Page 30 and 31: (If the velocities are not along th
- Page 32 and 33: KEY CONCEPTPhysicalChemistryFundame
- Page 34 and 35: ⎛ n ⎞ ⎛ n ⎞Correction term
- Page 36 and 37: Kinetic Isotope Effects :The kineti
- Page 38 and 39: (m / 32) 1Moles fraction of O 2 ==(
- Page 40 and 41: `tà{xÅtà|vtÄ V{tÄÄxÇzxá1 Se
- Page 42 and 43: Students' ForumExpert’s Solution
- Page 44 and 45: MATHSCOMPLEX NUMBERMathematics Fund
- Page 46 and 47: MATHSMATRICES &DETERMINANTSMathemat
- Page 48 and 49: aBased on New PatternIIT-JEE 2012Xt
- Page 50 and 51: dxO12. The time of crossing the riv
- Page 52 and 53: 6. Choose the INCORRECT statement f
- Page 54 and 55: 7. If the equation 3x 4 - 16x 3 + 3
- Page 56 and 57: Based on New PatternIIT-JEE 2013Xtr
- Page 58 and 59: 17. A stone is projected from level
- Page 60 and 61: 16. Half a mole of photon is used t
- Page 62 and 63: 17. Ifsin 3θcos 2θ2-147137= 0Then
- Page 64 and 65: IIT-JEE 2011PAPER-I (PAPER & SOLUTI
- Page 66 and 67: 9. Amongst the given options, the c
- Page 68 and 69: 22. The total number of alkenes pos
- Page 70 and 71: 1 2 × 8∆U = × [V - 0]22 2 + 81
- Page 72 and 73: shown in the figure. We use the sig
- Page 76 and 77: MATHEMATICSSECTION - ISingle Correc
- Page 78 and 79: Ans.Sol.[A,D]r = xiˆ + yj ˆ + zk
- Page 80 and 81: ⎡323 21× ⎤=⎡32 1⎤1 C2C21 C
- Page 82 and 83: XtraEdge for IIT-JEE 80 MAY 2011
- Page 84 and 85: Ans. [A]Sol. ∆T = k f × m × i
- Page 86 and 87: SECTION - IVMatrix match TypeThis s
- Page 88 and 89: Ans.[D]m 1 = 0.01 kgvH = 5mm 2 = 0.
- Page 90 and 91: 31. A series R-C circuit is connect
- Page 92 and 93: P3PBAColumn-I(A) Pipe closed at one
- Page 94 and 95: 21/ x45. If lim [1 + xl n(1 + b )]
- Page 96 and 97: So f(x) is monotonically decreasing
- Page 98 and 99: (C)25 / 6πI = x dxn ∫sec πl 327
- Page 100 and 101: Subscription Offer for Students'Xtr
- Page 102: XtraEdge for IIT-JEE 100 MAY 2011
43. Four solid spheres each of diameter 5 cm andmass 0.5 kg are placed with their centers at thecorners of a square of side 4 cm. The momentumof inertia of the system about the diagonal of thesquare is N × 10 –4 kg-m 2 , then N is.Ans. [9]Sol.mmmr =lm5 cm = × 1025 –2 m2m = 21 kgl = 4 × 10 –2 mUsing parallel axis theorem⎡ 2 1 5 −4 ⎤I total = ⎢4 × × × × 10 ⎥ +⎣ 5 2 4 ⎦⎡ 1 −4 ⎤⎢2× × 8×10 ⎥⎣ 2 ⎦⇒ 10 –4 + 8 × 10 –4 ⇒ 9 × 10 –4 kg m 244. The activity of a freshly prepared radioactivesample is 10 10 disintegrations per second, whosemean life is 10 9 s. The mass of an atom of thisradioisotope is 10 –25 kg. The mass (in mg) of theradioactive sample is.Ans. [1]Sol. A = λN10 10 1=9 × N10N = 10 19mass of sample = 10 19 × 10 –25 × 1 × 10 6 = 1 mg45. A long circular tube of length 10 m and radius 0.3carries a current I along its curved surface asshown. A wire-loop of resistance 0.005 ohm andof radius 0.1 m is placed inside the tube with itsaxis coinciding with the axis of the tube. Thecurrent varies as I = I 0 cos(300 t) where I 0 isconstant. If the magnetic moment of the loop is Nµ 0 I 0 sin(300 t), then 'N' is.Ans. [6]Sol. B =µ 0 0I cos300 t10µ Iφ = 0 0× 3.14 × 0.01 cos 300 t10φ = 3.14 × µ 0 I 0 cos 300 t × 10 –3dφe = − = 3.14 × 300 µ 0 I 0 sin 300t × 10 –3dtI−3e 3.14 × 300 µ I sin 300t 10i = =0 0 ×R 0. 005i = 3.14 × 60 µ 0 I 0 sin 300 t0.01Magnetic moment = 3.14 × 60 × 3.14 × × 100µ 0 I 0 sin 300 t= 5.9 µ 0 I 0 sin 300 t= 6 µ 0 I 0 sin 300 t46. Steel wire of length 'L' at 40ºC is suspended fromthe ceiling and then a mass 'm' is hung from itsfree end. The wire is cooled down from 40ºC to30ºC to region its original length 'L'. Thecoefficient of linear thermal expansion of the steelis 10 –5 /ºC, Young's modulus of steel is 10 11 N/m 2and radius of the wire is 1 mm. Assume that L >>diameter of the wire. Then the value of 'm' in kgis nearly.Ans. [3]Sol.mmgL∆L = = Lα(∆θ)AYAYα(∆θ)⇒ m ==gm = 3.14 kg ⇒ 3 kgπ × 10−611× 10 × 1010−5× 10XtraEdge for IIT-JEE 73 MAY <strong>2011</strong>
Change language