Paper-1-PART-II (PHYSICS)

PART-II (PHYSICS) 1. x  R 2 : If we displace the particle in forward direction, if m0g  so the particle will move away from equilibrium = unstable equilibrium. x  R 2 : If we displace the particle in forward direction, g = m0g  . So, particle will come back to equilibrium. So, stable equilibrium. 2. At t = 0  Displacement x = x1 + x2 = 4 sin 3 Resulting Amplitude A = Maximum speed = A = 20 m/s = 2 m.   = 2 m Maximum acceleration = A2 = 200 m/s2 1 Energy of the motion = 2 m2 A2 = 28 J Ans. 3. R2 – (R – 5)2 = (5 3 )2 R2 – R2 (R – 5)2 = (5 3 )2 R2 – R2 – 25 + 10 R = 75 R = 10 m x sin  = 1 , a = 30° ,  = 90 –  = 60° 2 mv = R  v = RqB qB m 10  106  10 = 5  105 = 2 m/s 4. For observer O ,  = V  Vs = V  V / 5  4V 1 1 f f 5f For O2, there is change of medium hence at the surface of water, keeping frequency unchanged  w = 4a 16V = 5f V  4V a w W 4V f = velo. of wave relative to observer w 4V  V = 5 w = 21V . 5 5 f 16V 21f = 16 5. max = (    I2 )2    I 2 2 < 4I min =    > 0  6. If unlike  dipole, E will monotonically decrease as we move away.  Like charge (can be both + or both –)  Direction of E at all points on the perpendicular bisector will be along the perpendicular bisector. There is a point at which E is maximum as derived below : 2Kq E = r 2 . cos  ; r = 2Kq a 2 sin   E = a2 . 4 sin2  cos  E is maximum when sin2  cos  is maximum. sin2  cos  = cos  – cos3  Differentiate and equate to zero. – sin  + 3 cos2  sin  = 0  sin  = 3 cos2  sin   sin  = 0  at infinity  minimum or 1 = 3 cos2   cos  =  tan  = a / 2 x = tan  =  P should be closer than . 7. Initial state is same for all three processes (say initial internal energy = E ) In the final state, V = V = V and P > P > P  P V > P V > P V  E > E > E if T > T then E > E for all three processes and hence (E – E ) < (E – E ) < (E – E )  E  < E  < E  If T < T , then E < E for all three processes and hence (E – E ) > (E – E ) > (E – E )  E  > E  > E  8. Magnetic field at centre (site of nucleus) B = 0  0qf = 0qv ]  B 1 and B  v 2r 2r  B  1 n5 2r  2r r 2 B1  (2)5 B2 (1)5 (Since, n1 = 1 to n2 = 2)  B1 = 32 B2 h h Also, mvr = n. 2 ,therefore angular momentum is changed by 2 . 9. Initially kx0 = (10 + 30) g compressed length x0 = 0.4 m = 40 cm Let the block of 30 Kg will reach height x above the natural length, then by energy conservation. 1 30g (0.4 + x) + 2 1 kx2 = 2 k (0.4)2 after solving x = – 0.2 m i.e. maximum height gained from initial position = 0.4 + x = 0.2 m = 20 cm 10. Let the block will reach height y above natural length, 1 1 then by energy conservation 10g (0.4 + y) + 2 After solving y = 0.2 i.e. maximum height gained from initial position = 0.4 + y = 0.6 cm = 60 cm . ky2 = 2 k (0.4)2 11. Let at an instant vR = (V ) cost  2 = 4 cost cost = ½  cost = 60°. Since VL is 90° ahead of VR vL = (VL)m cos (t + 90 ) = – 3 sint = – 3 sin 60° = – 3 cos 30°  | (V ) | = 3 cos 30° 12. From phasor diagram (VS)m = tan  = VL )m  3 = 5 volt.   = 37° VR )m 4  | v | = | (V ) cos (t + 37°)| S S m = 5 | cos (60° + 37°)| = 5 | cos 97°| = 5 cos 83° dB 13. dt = 2T/s E = – AdB = – 800 × 10–4 m2 × 2 = – 0.16 V dt 0.16 i = 1 = 0.16 A, clockwise dB 14. At t = 2s B = 4T; dt A = 20 × 30 cm2 = 600 × 10–4 m2; = 2T/s dA = –(5 × 20) cm2/s dt = – 100 × 10–4 m2/s E = – d d(BA)  = – = – BdA  AdB  dt  dt   dt dt  Alternative : = – [4×(–100 × 10–4) + 600 × 10–4 × 2] = – [–0.04 + 0.120] = – 0.08 v  = BA = 2t x 0.2 (0.4 – vt) = 0.16t – 0.4 vt2 d E = – dt at t = 2s t = 2s ij = 0.8 vt – 0.16 15. P = V is E = – 0.08 V where V = accelerating voltage is = saturation photocurrent Power of source of light  Quantum efficiency  (in Å) is = 12400 100 .01124 = 12400 = 0.1 A  Power = 100 watt 16. Power provided by accelerating potential = 100 W Power consumed by the source of 124 Å = 100 (0.5) = 200 W Overall efficiency = = .01 = 1% net power of X ray net power supplied 3W = 100W  200 W 17. (P)  Since focal length of the system is –ve so it will diverge the beam. (Q, R)  If refraction is through plane surface and incident medium is denser than refracting medium beam diameter increases and vice versa. (S)  since f > 0 so it will converge the beam 19. (P) PV = nRT P = (nRT) 1 = (constant) 1 V V T = constant i.e. isothermal process  V increases, W is positive and Q = U + U = W > 0 (Q) Q = 0 pdV = W = positive (R) PV = nRT As volume increases, T also increases i.e., U > 0 PdV = W > 0 So Q > 0 (S) For cyclic process U = 0 W < 0 (anticlockwise) Q < 0. 20. Let mx, my be nuclear masses of x and y P. mx – my – m = (Mx – zme ) – [My – (z – 2)me] – (MHe – 2me) Q. mx – my – m – me = (Mx – zme) – (My – (z–3) me) –(MHe – 2me) – me = M× – My – MHe–2me. R. mx – my – me = (Mx – zme ) – [My – (z – 1)me] – me = Mx – My – 2me S. (me + mx) – my = me + (Mx – zme) – [My – (z – 1)me] = Mx – My (K-capture).