10.PRACTICE TEST-2 (Solution-Paper-2)

1. a = 8 m PAPER - 2 PART I (PHYSICS) They meet when Q displace 8 × 3 m more than P displace  relative displacement = relative velocity × time. 8 × 3 = (10 – 2) t t = 3 sec Ans. 3 sec 2. Limiting friction between A & B = 60 N Limiting friction between B & C = 90 N Limiting friction between C & ground = 50 N Since limiting friction is least between C and ground, slipping will occur at first between C and ground. This will occur when F = 50 N. 3. From given graphs : ax = 3 t and a 4 y   3   1  v =  x 3 8 t2 + C At t = 0, vx = – 3  C = – 3 3  3 2   v = t2 – 3  dx =  8  t  3 dt  .... (1)  3 t2  t   Similarly dy =   8 4 dt  .... (2) As dw = F. ds = F.(dx ˆi  dy ˆj) W 4  3 ˆ  3  ˆ  3 2  ˆ  3 2  ˆ  dw    4 t i   4 t  1 j .  8 t  3  i   t 8  t  4  j dt 0 0           W = 10 J Alternate Solution : Area of the graph ; ax dt = 6 = V(x) f  (3)  V = 3. and ay dt = –10 = V(y) f  (4)  V = – 6. Now work done =  KE = 10 J 4. The rate of heat loss by a thin hollow sphere of thickness ‘x’ , mean radius ‘r’ and made of density ‘’ is given by mS dT = –  A(T4 – T 4 ) dt 0 ( 4r2x) S dT =    4r2 (T4 – T 4 ) dt 0 dT  (T 4  T 4 )  dt = – 0 Sx is independent of radius Hence rate of cooling is same for both spheres.  dv  f0  𝑙  𝑙 5.   v  = max f0 𝑙 2  𝑙1 1 0.1 0.1  1  1  = 100 + 74  24 =  100 50   × 100% = 1.2% 2k 6. Electric field between the two cylinders = r  Force on charge q = This force is centrepetal force 2kq r  2kq r mv2 = r  v = = 7. Power maximum when r = R. So, power consumed by it will decrease for R > r. R v 8. v = – 30, m = – u = – 2  AB = CD = 2 × 1 = 2 mm Now BC BC = AD = v AD u2 = 4  BC = AD = 4 mm  length = 2 + 2 + 4 + 4 = 12 mm Ans. 9. 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 90 10. Time of flight, T = 15 = 6 sec. 120 gT  u = = 20 m/s, u = 15 m/s & u = = 30 m/s y 6 x z 2  velocity of proejction = ( 15ˆi  20ˆj  30kˆ )m/s. 11. L =  = constant  d  d = 0  d d = –     L2 K.E. = K = 2 dK d  = – K  As temperature increases,   ,  K  &   . 12. x = ( – 1) t = n  t = n (  1) n = 1, 2, 3, 4  t = 1.18, 2.36, 3.54 and 4.72 m 13. i = isteel + icopper 12   (0.5)2  104  100 = + 92 ×  × (12 – 0.52) × 10–4 × 100 2 = 150  × 10–4 + 3450  × 10–4 = 3600  × 10–4  Lsteel L 150 10 4 = 3600 10 4 = 0.04 = 4%. 14. Potential difference across AC is zero 5 – 2 = 0  = 2.5 A Let resistance of part BC = R Using KVL 10 + 5 – 2 – R –  = 0 R = 3. 15. 16. T2  R3 R  T2/3 So, radius of planet B is R ( 2 2 )2/3 = 2 times that of planet A  Radius of A is half that of B. 1 Since   R2 So intensity of sunrays at planet A to that at B is   1 (1/ 2)2 = 4 times. 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 18. From diagram Watt R H = 2 tan H 1 4 = R = 2 tan = 4 = 2 2 19. (A) F = constant and → u F  0 Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle will move in straight line and speed may increase or decrease. (B) → F  0 and F = constant initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing. (C) → F  0 means instantaneous velocity is alway perpendicular to force. Hence the speed will remain constant. And also | F | = constant. Since the particle moves in one plane, the resulting motion has to be circular. (D) →  2 ˆi  3ˆj and →  ˆ  9ˆj . Hence initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed. 20. (A) Electrostatic potential energy = 1 4 0 (Q)2 2a Q2 = 8 0 a 1 (Q) (Q)  (Q)2  3 Q2 (B) Electrostatic potential energy = 4   5a / 2 2(5a / 2)  = 20   a 0   0 1 3Q2 3 Q2 (C) Electrostatic potential energy = 4 0 5a = 20  0 a 1  3Q2  (Q)2  (Q) (Q)  27Q2 (D) Electrostatic potential energy = 4   5a 2(2a) 2a  = 80  a 0   0

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