[1] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Direction : An initially parallel cylindrical beam travels in a medium of refractive index (I) = 0 + 2I, where 0 and 2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. Q.1 The initial shape of the wavefront of the beam is -(1) planar (2) convex (3) concave (4) convex near the axis and concave near the periphery -Ans. (1) Shape of wavefront is planer for parallel beam of light. Q.2 The speed of light in the medium is -(1) maximum on the axis of the beam (2) minimum on the axis of the beam (3) the same everywhre in the beam (4) directly proportional to the intensity I Ans. (2) 1 I 1r r So, speed is minimum at axis and maximum at periphery. Q.3 As the beam enters the medium, it will (1) travel as a cylindrical beam (2) diverge (3) converge (4) diverge near the axis and converge near the periphery Ans. (1) As parallel beam is passing through a medium, normally so their direction do not change.[2] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.4 The speed of daughter nuclei is -(1) m c M m (2) m c M m (3) 2 m c M (4) m c M Ans. (3) M/2 M/2 E = mc2 K.E = 2 × 12 × M2 . v2 or, mc2 = 2 M v 2 or, v = 2 2 mc = c 2 m M Q.5 The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then (1) E1 = 2E2 (2) E2 = 2E1 (3) E1 > E2 (4) E2 > E1 Ans. (4) As energy is released in this fission due to mass defect, that is why daughter nucleus is more stable, so their binding energy per nucleon is higher i.e, E2 > E1 Q.6 When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by X-rays, both, V0 and Kmax increase. (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true.[3] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Ans. (1) As energy of X-ray is higher than that of UV-ray, that is why Kmax & V0 increases, as eV0 = Kmax = h – . So statement-1 is true. Kinetic energy of emitted electron vary between O to Kmax depending upon the layer from which they are emitted, as well as due to frequency of incident photon. So, statement-2 is partially true. But it do not fully explain Q.7 Statement Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true. Ans. (2) When both particles are moving in same direction then their total initial momentum can never be zero and so their final momentum after collission will also not be zero. Thus statement-1 is true. Statement-2 is also true and it correctly explains statement-1. Q.8 The figure shows the position-time (x – t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is -(1) 0.2 Ns (2) 0.4 Ns (3) 0.8 Ns (4) 1.6 Ns Ans. (3) v = dx 2m dt 2s = 1m/s v = 2 m/s Impulse= p = mv = 0.4 kg × 2m/s = 0.8 Ns[4] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.9 Two long parallel wires are at a distance 2d apart. They carry steady equal currects flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX’ is given by-(1) (2) (3) (4) Ans. (2) Variation of B with r for a current carrying wire is B = 0 0 I ˆr r 0 2r I ˆ – r r 0 2r Q.10 A ball is made of a material of density whre oil < < water with oil and water representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibriiu in a mixture of this oil and water, which of the following pictures represents its equilibrium position? (1) (2) (3) (4) Ans. (3) As, water > > oil so, water will be at lower portion, ball at middle and oil at top portion.[5] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.11 A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E at the centre O is -(1) 2 2 0 q ˆj 2 r (2) 2 2 0 q ˆj 4 r (3) 2 2 0 q ˆ – j 4 r (4) 2 2 0 q ˆ – j 2 r Ans. (4) = qr dE + + + + + + + + + + + + + + + + + + + + i j dE = o 1 4 2 dl r dEx = dE cos = 0 1 4 2 .rd .cos r dEx = 0 1 4 r cos . d d d r d rd Ex = /2 0 – /2 1 . r 4 . cos d = 2 0 q 1 r 4 /2 – /2 sin = 2 2 0 q 2 r xˆ E –E j 2 2 0 q ˆ E – j 2 r [6] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.12 A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of gthe gas increases from V to 32 V, the efficiency of the engine is -(1) 0.25 (2) 0.5 (3) 0.75 (4) 0.99 Ans. (3) V1 = V V2 = 32V For diatomic gas = 1 + 2f = 1 + 25 = 75 –1 –1 1 1 2 2 T V T V 7 1 –1 5 1 2 2 1 T V 32V T V V = (32)2/5 = 2 5 5 2 = 22 = 4 = 1 – 21 T 1 3 1 T 4 4 = 0.75 Q.13 The respective number of significant figures for the number 23.023, 0.0003 and 2.1 × 10–3 are-(1) 4, 4, 2 (2) 5, 1, 2 (3) 5, 1, 5 (4) 5. 5. 2 Ans. (2) Significants figures are 5, 1, 2 respectively. Q.14 The combination of gates shown below yields -(1) NAND gate (2) OR gate (3) NOT gate (4) XOR gate Ans. (2) X = (A.A) . (B.B) Truth Table = A . B A B A B Y 0 0 1 1 0 0 1 1 0 1 1 0 0 1 1 1 1 0 0 1 = A B = A + B = OR gate.[7] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.15 If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of teh spectrum called (1) -rays (2) X-rays (3) ultraviolet rays (4) microwaves Ans. (2) Power = 4kw = 4 × 103 J/s Produces No. of photon /sec = 1020 nhc = 4 × 103 1020 × –34 8 6.67 10 3 10 = 4 × 103 = –9 20 3 10 3 4 –9 5 10 m = 50Å X-ray Q.16 A radioactive nucleus (initial mass number A and atomic Z) emits 3 -particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be -(1) A Z 4 Z 2 (2) A Z 8 Z 4 (3) A Z 4 Z 8 (4) A Z 12 Z 4 Ans. (3) 0 1 A 12 A 12 (–2 ) –3 A ' Z 6 Z 8 Z X Y X No of neutrons = (A – 12) – (Z – 8) = A – Z – 4 No of proton = Z – 8 No of neutron A Z 4 No of proton Z 8 Q.17 Let there be a spherically symmetric. large distribution with charge density varying as (r) 0 5 r – 4 R upto r = R and p(r) = 0 for r < R is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by (1) 00r 5 r – 3 4 R (2) 0 0 4 r 5 r – 3 3 R (3) 00r 5 r – 3 3 R (4) 00 5 r 5 r – 3 4 R [8] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Ans. (3) = 0 5 r – 4 R r dr dq = 4r2dr 5 r – 4 R 0 Q(enclosed) = Q = r 2 0 0 5 r 4 r – dr 4 R = 40 r 3 4 0 5 r r – 4 3 4R = 40 4 3 5 r r – 12 4R = 4or3 5 r – 3 R = 0 Q = 3 00r 5 r – 3 R = E × 4r2 0 0r 5 r E – 4 3 R Q.18 In a series LCR circuit R = 200 and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circut the corrent lags behind the voltage by 30º. On taking out the inductor from the circuit the currect leads the voltage by 30º. The power dissipaate in the LCR circuit is -(1) 242 W (2) 306 W (3) 210 W (4) Zero W Ans. (1) In LCR R = 200 V = 220 f = 50 Hz Step I After taking out capacitance c then become L-R tan 30º = L L X R X R 3 [9] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Step II After taking out incluetance. then it become L-C tan 30º = C C X R X R 3 Step III In L-C-R Z = 2 2 C L (X – X ) R = R P = Irms Vrms cos = 2 rms VR cos = 220 220 200 × cos 0º P 242W Q.19 In the circuit shown below, the key K is closed at t = 0. The current through the battery is -(1) 1 2 1 2 V(R R ) R R at t = 0 and 2 V R = t = (2) 1 2 2 2 1 2 VR R R R at t = 0 and 2 V R at t = (3) 2 V R at t = 0 and 1 2 1 2 V(R R ) R R at t = (4) 2 V R at t = 0 1 2 2 2 1 2 VR R R R at t = Ans. (3) Step I at t = 0 time when k is closed XL = flowing current in battery 2 V I R Step II at t = time total ristance of circuit R’ = 1 2 1 2 R R R R I’ = V R ' = 1 2 1 2 V(R R ) R .R[10] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.20 A particle is traving with volocity ˆ ˆ K(yi xj) , where K is a’ constant. The general equation for its path is -(1) y2 = x2 + constnat (2) y = x2 constant (3) y2 = x + constant (4) xy = constant Ans. (1) vx = ky and vy = kx x dv kdy dt dt ax = k2x or x x v dv dx = k2x 2x v2 = 2 2 k x c 2 vx2 = k2x2 + 2c k2y2 = k2x2 + 2c y2 = x2 + const Q.21 Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for teh charge to reduce to one-fourth its initial value. Then the ratio t1 /t2 be -(1) 2 (2) 1 (3) 12 (4) 14 Ans. (4) Q = Qoe–t/RC t = RC ln 0 QQ Ui = 2 0 Q 12 C Uf = i U2 Q = 0 Q2[11] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 t1 = RC ln 00 Q 2 Q t1 = 12 RC ln 2 for charge varition t2 = RC ln 0o Q 4 Q = 2RC ln2 11 1 RC n 2 t 2 t 2RC n 2 l l = 14 Q.22 A retangular loop has a sliding connector PQ of length l and R and it is moving with a speed as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are -(1) I1 + I2 = Bl 6R , I = Bl 3R (2) I1 + –I2 = Bl R , I = 2Bl R (3) I1 + I2 = Bl 3R , I = 2Bl R (4) I1 + I2 = I = Bl R Ans. (3) I = E B v 2B v Re q R /2 R 3R I1 = I2 = B v 3R[12] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.23 The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y -0.02(m)sin t x 2 – 0.04(s) 0.50(m) . The tension in the string is -(1) 6.25 N (2) 4.0 N (3) 12.5 N (4) 0.5 N Ans. (1) Hint v = T Q.24 Two fixed frictionless inclined planes making an angle 30º and 60º with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B -(1) 4.9 ms–2 in vertical direction. (2) 4.9 ms–2 in horizontal direction (3) 9.8 ms–2 in vertical direction (4) Zero Ans. (1) g sin 60º 60º g sin 30º 30º avertical = g sin 60º × cos30º a'vertical = g sin 30º × cos 60º = g × 3 3 2 2 = g × 12 × 12 = 3g 4 = g4 vA vB a – a = 3g 4 – g4 = g2 = 4.9 ms–2[13] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.25 For a particle in uniform circular motion the acceleration a at a point P(R, ) on the circle of radius R is (Here is measured from the x-axis) (1) 2 v ˆiR + 2 v ˆj R (2) – 2 v ˆ cos i R + 2 v ˆ sin j R (3) – 2 v ˆ sin i R + 2 v ˆ cos j R (4) – 2 v ˆ cos i R – 2 v ˆ sin j R Ans. (4) V2 R V2 R V2 R v a = c v2 R i a = – 2 v ˆ cos i R – 2 v ˆ sin j R Q.26 A small particle of mass m is projected at an angle with the x-axis with an initial velocity v0 in the x-y plane as shown in the figure. At a time t < 0 sin g , the angular momentus of the particle is -(1) 12 mg 0 cos ˆi (2) – mg t2 cos ˆj (3) mg 0 t cos ˆk (4) – 12 mg 0 t2 cos ˆk where ˆ ˆ i, j and ˆk are unit vectors along x, y and z-axis respectively.[14] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Ans. (4) At t < 0 v sin g particle does not orose manimum ereight x = uxt = (ucos .t) ˆj and y = uyt – 12 gt2 = usin t – 12 gt2 r = u cos t i + (usint – 12 gt2) j Now, vx = ux = ucos vy = ux = – gt = usin – gt v = u cos i + (u sin – gt) j L = r p m(r v) = m[ucost i + (u sin t – 12 gt2) j ] × [u cos i + (u sin – gt) j ] = – 12 mug cos .t2 ˆk Q.27 Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each. suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is -(1) 1 (2) 4 (3) 3 (4) 2 Ans. (4) mg = T cos . F = T sin F = mg tan 0 1 4 1 2 2 q q = Vg tan [15] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 In liquid, mg – U = T' cos . F' = T' sin F' = (mg – U) tan 0 r 1 4 1 2 2 q q = Vg tan (1) (2) r = = 1.6 2 1.6 0.8 Q.28 A point P moves in coungter-clockwise direction on a circular path as shown in the figure. The movemeen of ‘P’ is such that is sweeps out a length s = t3 + 5, where s is in metres ant t is in second. The radius of the path is 20 m. The acceleration of ‘P’ when t = 2 s is nearly-(1) 14 ms2 (2) 13 m/s2 (3) 12 m/s2 (4) 7.2 m/s2 Ans. (1) v = 2 ds 3t dt v is tangeutial to s. ar = 2 2 v at R R again, dv 6t dt at t = 2 sec at = 36 5 m/s2 = 7.2 m/s2 and aT = 6 × 2 = 12 m/s2 a = 2 2 r t a a = 14 m/s2[16] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.29 The potential energy fuction for the force between two atoms in a diatomic molecule is approximately given by U(x) = 12 6 a b – x x , where a and b are constants and x is the distance between the atoms. If the dissociation energy of teh molecule is D = [U(x = ) – Uat equilibrium], D is -(1) 2 b6a (2) 2 b2a (3) 2 b 12a (4) 2 b4a Ans. (4) For equilibrium. dU 0 dx 13 7 –12a 6b – 0 x x 6 2a – b 0 x x6 = 2a b x = 16 2a b Uat eqb = 2 a 2a b = – b 20 b = 2 2 b b – 4a 2a = – 2 b4a and U = 0 D = 2 b4a [17] Regd. Office : 638, CAD circle, Dadabari Main road, Kota (Raj.) Ph. 0744-2502409 SOLUTIONS OF AIEEE -2010 PHYSICS PAPER 0744-2502409, 09414187273 Q.30 Two conductors have the same resistance at 0ºC but their temperature coefficients of resisdance are 1 and 2. The respective temperature coefficients of their series and parallel combinations are nearly -(1) 1 2 2 , 1 2 2 (2) 1 2 2 , 1 + 2 (3) 1 + 2, 1 2 2 (4) 1 + 2, 1 2 1 2 Ans. (1) R'1 = R1 (1 + 1T) and R'2 = R2(1 + 2 T) R'series = 1 1 1 2 R R Rs (1 + s T) = R1 (1 + 1 T) + R2(1 + 2 T) (R1 + R2) (1 + 3T) = (R1 + R2) + (R11 + R22)T s = 1 1 2 2 1 2 R R R R = 1 2 2 and for parallel ' ' ' p 1 2 1 1 1 , R R R p p 1 R (1 T) = 1 1 1 R (1 T) + 2 2 1 R (1 T) 1 2 p (1 – T) (1 – T) 2 (1 – T) R R R p = 1 2 2