Paper-2-PART-I (PHYSICS)

PART-I (PHYSICS) mg 1. The initial extension in spring is x0 = k Just after collision of B with A the speed of combined mass is v . 2 For the spring to just attain natural length the combined mass must rise up by x0 = comes to rest. mg(sec fig.) and k Applying conservation of energy between initial and final states 1  v 2 1  mg 2  mg  2 2m  2  + k   2 k = 2mg  k        Solving we get v = Alternative solution by SHM  ] 1  1 2. = (1.5 – 1) 1  1 = Flens F𝑙 = 80 cm   40  20  80 Fm = – 20 = – 10 cm 2 1 Feq  1  2 fm f𝑙 1 =  10 2 – 80 feq = – 8 cm Hence object should be placed at x = 16 cm, i.e. at the centre of curvature. 3. Averaging a number of readings makes the measurement more precise. 4. x = x1 and x = x3 dU are not equilibrium positions because dx 0 at these points. x = x2 is unstable, as U is maximum at this point. 5. Tension in both string shall be same which can be observed by making FBD of string in figure (1) 6. Let the length of rod be ‘a’. The magnitude of transverse magnification of ends A and B is 2 each. The image of B is virtual of A is real. Applying mirror formula to B and A 1   a   1   1 a  f .... (1) 2 L   2 L   2     1   a    1   1 a  f .... (2) 2L  a    2L     L     Solving we get a = L. 7. The slope of temperature variation is more in inner dQ = dt KA .T 𝑙 T = 𝑙 . dQ KA dt Slope  1 K Larger the conductivity, smaller is the slope. 1 2 hc 8. mv   2 Clearly 1 m2 2 v'  hc = (3 / 4) v –  = 4hc   3 9. There can be three minima from central point to  corresponding to  , 3 , 5 path differences.  total number of minima = 2nmax = 6 10. Law of equipartition of energy 2 2 2 < KER > = 2 . 1 kT = E 2  < KE > = 3. 1 kT 2 3E < KET > = 2 11. Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside. Field due to only q is non-zero. 12. Speed of wave in wire V = = = Maximum time period means minimum frequency ; that means fundamental mode. V f =  V = 2𝑙  T = 2𝑙 = 2𝑙 1 = 35 second 𝑙 =   (f = 35 Hz) and; frequency of first overtone = V = 70 Hz. 𝑙 13. In the initial state, charge on each capacitor is shown in figure-1. Let charge q flow anticlockwise in the circuit before it achieves steady state as shown in figure-2. Applying KVL to figure 2. 360  q  3 q 1.5  300  q  0 2 or 180 C  final charge on 1.5 F capacitor is q = 180 C and final charge on 2 F capacitor is 300 – q = 120 C. 14. 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 15. For case–1 : gh = v1 = 1 v 2 2 t1 = For case–2 : g2h + 2gh = 1 zv 2 2 v2 =  t = For case–3 : g2h + 2g2h + 3gh = v3 = 1 3v 2 2 t3 = x1 = v1t1 = x1 : x2 : x3 = = = x2 = : : : : : : . x3 = 16. aA = a =  . R ...(i) T – mg = 0 ...(ii) mR2 T.R = . (iii) 2  g = a 2 acm = 0 17. Area perpendicular to the light = 1 x cos600 Energy falling on the surface=intesity x perp.area =3 x 1 x cos600 = 3 2 Momentum carried by the light per sec = 3/(2c) = 5 x 10-9 Watt 18. V0 sin30° = V1 cos30° (i) eV0 cos30° = V1 sin30° (ii) Dividing (i) & (ii) 1 e tan 30° = 1 tan 30 1 e = 3 → → 19. W = F.dr =  ydx  xydy   2x = 3y 2  3  0  ydy  3 y 2dy  2   dx = 3 dy 2  3 =  4 y2  y3 2  0 = 7 Joule. A  1 t /  20. 1    A0  2  100 / 20  1 t /  t 1   141/ 20    2    2   = 2t = 2 × 3 days = 6 days.  