Paper-2-PART- II (PHYSICS)

PART- II (PHYSICS) 1. (A) It will fall because mg is acting on it towards the centre of planet and initial velocity is zero. It'll move in straight line. (C) Time of fall can be found by two methods : I Method : By energy conservation 1 mv2 – 2 GMm = 0 – r GMm – (1) R using this we get V = f(r). Now use dr V = – dt R' dr  f (r) = – dr dt t     f (r)  dt 0 ; R' = radius of the planet. In the final expression (or in the beginning itself) R'  0 { R > > R'} you will get t = Here GMm R 2  2 2 = m T  R   Note : This method is longer. If a student gets idea of solving the question only by this method then it is better to leave this question because it will consume more time. II Method : Kepler's Law : T2  r3 . Assume that the satellite moves in elliptical path with maximum and minimum distances from centre as R and R'.  R > > R'  velocity at R is very small ( ~ 0). When it reaches R' then it touches the surface of the planet. This motion (from R to R') is almost, same as given in the question. T2 r 3 Now 1  1 , T 2 2 R  R R r2 = 2 ~ 2 = T, r1 = R  T2 = 2.  = = 10 rad/s T = 2  2 s  10 2  Maximum speed will be at the natural length of the spring T/4 = 10  4 = 20s. Time taken to cover 0.1 m is T   s 4 20 Time taken to cover 1 × 0.1m is T  2 = 2  2   s 2 4 3 10  4 3 30 3. Portion in x–y plane will not experience any net force due to x–component of magnetic field. Only its z–component will be effective. So, current in the loop is as shown. → R2 ˆ ˆ  M  0(k  j) 2 net force on the portion is in x-z plane = 0  002 ˆ 2Rˆi  M00 (ˆi  kˆ) 2R =  j → → 002R ˆ ˆ ˆ Net torque = M B = (i  j  k) 4 Net magnetic field at the origin = 00 (ˆj  kˆ)  00 (ˆi  kˆ) . 4R 2R 4. fobs = f [vs  v cos ] [vs  v cos] = f obs = vs  v cos  f For any observer in train frequency observed is equal to original frequency but observed wavelength is more. 5. x = ( – 1) t = n  t = n (  1) n = 1, 2, 3, 4  t = 1.18, 2.36, 3.54 and 4.72 m 6. Potential at each point on y-z plane is zero. The electric field will be zero on y-z plane at a distance 2 a from origin. n1CV  n2CV 7. CV = n1  n2 C eq and CPeq  C  R ;  = CVeq 8. For Balmer series, n1 = 2, n2 = 3,4,........ (lower) (higher)  In transition (V), Photon of Balmer series is absorbed. In transition  E2 = – 3.4 eV, E4 = – 0.85 eV E = 2.55 eV E = hc   = hc  E  = 486 nm. Wavelength of radiation = 103 nm = 1030 Å  E = 12400 1030 Å ~– 12.0 eV So difference of energy should be 12.0 eV (approx) Hence n1 = 1 and n2 = 3 (–13.6)eV (–1.51)eV  Transition is V. For longest wavelength, energy difference should be minimum. So in visible portion of hydrogen atom, minimum energy emitted is in transition V. 9. At the instant block is in equilibrium position, its speed is maximum and compression in spring is x given by kx = mg (1) From conservation of energy kx = mg (1) mg (L + x) = 1 2 from (1) and (2) we get vmax = kx2 + 1 mv2 2 . max .... (2) 10. V max = and  = = 2  A = Vmax 3  = 4 L Hence time taken t, from start of compression till block reaches mean position is given by x = A sin  t where x = L 4 1  t = sin–1 3 2 Time taken by block to reach from mean position to bottom most position is 4 = 1 Hence the required time = + sin–1 3 11. As current is leading the source voltage, so circuit should be capacitive in nature and as phase differ-  ence isnot 2 , it must contain resistor also. 12. Time delay =    = 400 Þ    4  1   1 tan–1  RC  = Þ = R   4 C i0 = V0 100 =  R = 50  and C = 1 50 100 = 200 µF p 13. r = qB > d  p > qBd = 2eBd for  particle p 14. r = qB < d  p < qBd = eBd for proton/electron (neutrons pass over to region  undeflected) 15. As Rate of decay = Rate of production  P = N  N = P  = 16. As N = Pt1/ 2 𝑙n2 Pt1/ 2 𝑙n2 = 1.8 × 1015 it is dependent upon P and t 1/2 . Initial no. of 56Mn nuclei will make no difference as in equilibrium rate of production equals rate of decay. Large initial no. will only make equilibrium come sooner. 17. y = y + 4 y  dyEF = dy2 + 4 dy1 EF 2 3 1 dt VEF dt = 6 + 3 4 (–4) 3 dt = 6 – 16 = 3 18 – 16 3 = 2 cm/s 3 y = y + 3y2  dyfE = dy1 + 3 dy2 FE 1 4 dt dt = (–4) + 4 dt 3 (6) 4 = –4 + 9 = 2 –89 = 2 1 cm/s 2 y = y1 +y2  dy = dt dy1 + dt dy2 = –4 + 6 = 2cm/s dt dy1 = –4 cm/s dt 18. x = t – 2 t2 + 3 + 5 3 v = t2 – 4t + 3 = t2 – 3t – t + 3 = (t –1) (t – 3) a = 2t – 4 = 2 (t – 2) 19. A-B isothermal U = 0 W = RT0 ln2 Q = W = RT0 ln2 B-C isochoric U = 3 RT 2 0 W = 0 Q = U = 3 RT 2 0 C -D isothermal U = 0 W= –2RT0ln2 Q = W =–2RT0ln2 D-A isochoric U = –3 RT 2 0 W = 0 Q = U = –3RT0 2 20. (P) Stopping potential in electron volts = hv–= 12–4 = 8.  Z  1 2   85 2 (Q)  1   2 =   . Therefore Z = 86 and Z =82  Z2  1 1  81  1 2 (R) Half life time of radioactive material is 4 min. For 160 gm to reduce to 20gm, Three half life times are required. (S) The binding energy per nucleon for helium in MeV is approximately 0.0302  930  7 4