Paper-3-Part Test-I

1. KE A/CM = 1 .1.(v 2 PART-I (PHYSICS) )2 = 2 Joules A/CM = 2 m/s. Let ; COM move towards +ve x-direction. Then, → v A / CM 2ˆi → → B/CM =  ˆi ( Use ; → vCM m1v1  m2 v2 m1  m2 )  KE = 1 .1.2ˆi  → 2 + 1 .2 ˆi  → 2 System 2 v CM 2 vCM = 1 4  v2  ˆ →  1 .2.1 v 2  ˆ →  2 CM 2.2i.v CM 2 CM 2.i.v CM = (2  2  → 2i.vCM )  (1 4  → 2i.vCM ) = 9 J 2. Focal length is minimum in case I, therefore power is maximum. 8 3. Least count = 1 – 10 = 0.2 mm Length = 4 + 5 × 0.2 = 5.0 mm = 5.0 × 10–3 m. → 4. f U U = – x ˆi – y → ˆ ˆ ˆj = – [6 ˆi ] + [8] ˆj = – 6 ˆi + 8 ˆj →  3ˆi  4ˆj  →   a = – 3 i + 4 j has same direction as that of u    a  2 2   | → | = 5 , | → | = 5/2 a u Since → and → are in same direction, particle will move along a straight line u a  S = 5 × 2 + 2 1 × 5 × 22 = 5 + 10 = 15 m. 15 m. Ans 2 0 𝑙 m 2 5. – T   dx2x  T = m 2 x T 0 𝑙 𝑙 2  Y = F𝑙 A𝑙 𝑙 = m 2 x2 F𝑙 Ay dx m 2𝑙3 2𝑙3 𝑙 = 𝑙 2 AY 𝑙 = 𝑙 6AY 𝑙 = 6y 𝑙 = 2 = 2 6. Clearly, PM = 3 cm 2 37º > sin–1 0 1  a(3 / 2) 3 5 > 3n + 1 3a n0 2 9a > 5 0 2 9a > 1 2 2 a > 9 7. P = AeT 4 2 = 2 x 10-6 x0.9x 5.6 x 10-8 x T4 1014 T4 = 0.9 x 5.6 T = 2110 k 8. As the accelerating potential difference is changed only the minimum wavelength changes. It has no effect on wavelengths of characteristic x-rays (whether they are produced or not)  (C) is the correct choice. 9. (B)  = 2t 4 = 2 × 3 8 × 1 = 3 There are two phase reversal of . One at the air water surface 8 another at the water-ground surface.  condition for maximum interference will be  = n 8 3 = n  = 8  = 8 m 3n max 3 10. In an adiabatic expansion, internal energy decreases and hence temperature decreases.  from equation of state of ideal gas PV = nRT  The product of P and V decreases. 11. V at origin  0 K(q)r  K.Q.r  1   E(r = 2 m) = (R2  r 2 )3 / 2 (R2  r 2 )3 / 2 = K.rq   103 / 2 23 / 2.103 / 2  = 0 From origin to r = 2, field is towards origin. 12. The mechanical strain  𝑙 = 𝑙 =  T = 1.21 × 10 × 20 = 2.42 × 10 –5 –5 The tension in wire  𝑙 = T = Y 𝑙 A = 2 × 10 × 2.42 × 10 × 10 = 48.4 N 11 –5 –6  speed of wave in wire V = Since the wire is plucked at 𝑙 4 = from one end = 22 m/s The wire shall oscillate in 1st overtone (for minimum number of loops)  = 𝑙 = 1m V Now V = f  or f =  13. Ex = 3x2 + 0.4 N/C = 22 Hz. V = E dx = (3x2  0.4)dx =  3 0.2 0.2  x 0 x  0.4x 0 V = (0.2)3 + 0.4 × 0.2 = 0.088 volt C = Q = V 0.88 0.088 = 10 µF 14. For given condition : Magnitude of Bsolienoid = Magnitude of Bloop  ni = 0 2R here n = Total no. of turn Total length 1300 = 0.65  1 i = 2R  n 8  0.65 = 2  0.02  1300 = 100 mA. For given condition : Total magnetic field at the centre of loop = |Bloop| + |Bsolenoid |  |Bloop| = ||Bsolenoid| = 2|Bloop | = 2 × 0  2R 2  4 107  8 = 2  0.02 = 16  × 10–5 T. 15. The pressure at any point can never have different values. Hence (A) & (D) are not possible. (Calculate the pressures at points A & D from both their left and right) In case of insufficient length of capillary tube the shape of meniscus is as below : 16. Minimum velocity required at D  v = mv 2 R  r = mg Energy conservation between A and D mg(h – 2R + r) = 1 mv2 + 1 mr 2 v 2 × 2 g(h – 2R + r) = 4h  7r 2 2 2 r 3 g(R – r) 4 52  14 R = 11 17. By moseley's law, = 11 = 6 cm. = a(z – b) z1  1 = z2  1 z1  1 2 = z2  1 36 2 = z2  1 z2 – 1 = 18 z = 19 = 95/n  n = 5 18. Ndt = mNdt = m. (i) & (ii)  (i) .............(ii) 19 .i = f 2  = 3 mg L sin + 0 = m g (L  x) sin – mg x + 1 mv2 2 1 2 2 2 L g x2 mg 2 sin = L (L2 + x2 – 2Lx)sin – + v2 g x2 gL sin = L v = (L2 + x2 – 2Lx) sin  – g + v2 L A  1 t /  20. 1    A0  2  100 / 20  1 t /  t 1   141/ 20    2    2     = 2t = 2 × 3 days = 6 days.