FOR XEROX PHYSICS

SOLUTION of DPP no. 27 Take g = 10 m/s2 whereever required in this paper. Q.1 (a) N, (b) (i) Y, (ii) N Q.2 Dimensional formula for K = ML2T–2 ;  = T–1 ; r = L K = 1 X 2r2 2 Using dimensional homogeneity ML2T–2 = X [T–1]2 (L)2  [X] = M X represents mass Q.3 F1 = 10 cos 53 ˆi +10 sin 53 ˆj = 6ˆi + 8ˆj F2 = 10 cos(37) ˆi +10 sin (37) ˆj = 10 cos 37 ˆi 10 sin 37 ˆj = 8 ˆi  6 ˆj (N) Fmt = F1 + F2 = 14 ˆi + 2ˆj  Fmt ˆ ˆ a = = (14i + 2 j) m/s m Q.4 geff = g – a = 8 Mt/s2 SB, L = 2 metre downward  is positive uB, L = 0 1 2 = 2 8 t2  t = sec Q.5 (a) Vp, g = Vp, sidewalk + Vsidewalk , g 80 = 4iˆ +1iˆ = 5iˆ m/s t = 5 s = 16 s (b) While moving in opposite direction Vp, g = 3iˆ m/s ; t = – 80 – 3 = 26.67 Total time (T) = 16 + 26.67 = 42.67 Q.6 Let acceleration be a. Man is late by T and length of each boggy is L. Velocity of train when man enters = 0 + aT = aT Considering time taken to cross IInd last boggy L = (aT)3 + 1 a(3)2 (1) 2 Considering time taken to cross last boggy L = a (T+3)2 +  L  1 a(2)2 (2) 2 Equating  a  from equation (1) and (2)   9 3T + 2 7 T = 2 = 2T + 6 + 2 ; sec. Ans. Q.7 V = 3t 2 iˆ + 3 ˆj  r  t   dr =  v dt r0 0 t t  r  r0 =  3t 2 dt iˆ +  3 dt ˆj = t 3 iˆ + 3t ˆj 0 0   = + t i + 3t j At t = 2 sec. r r0  ˆ ˆ ˆ = ˆ ˆ r = 2 j + 8i + 6 j For average acceleration Initial velocity at t = 0 8i + 8 j Vi = 3 ˆj Final velocity at t = 2 sec. Vf = 12 iˆ + 3 ˆj a = Vf  Vi t 12 iˆ = 2 = 6 iˆ Ans. Q.8 (a) t2 – 5t + 6 = 0 (t – 2) (t – 3) = 0 t = 2, 3 s (b) at t = 0, x = 6, v = 2t – 5 5 5 v = 0 at t = 2 5 1 v > 0 for t > 2 distance travelled = 6.5 mts. at t = 2 , x = – 4 Q.9 x = 2 + area of curve v–t t 0 1 2 3 4 5 6 x 2 4 6 7 6 4 2 Q.10 ux' = u cos 30° = u uy' = u sin 30° = 2 ax' = – g sin 30° ay' = – g cos 30° 2u 1  2uy'  2  T = =   = 2 g cos  10 3 2  u = 10 m/s R = x' = u 1 1 t + a t2 = 15 × 2 + (–5) × (2)2 = 20 m Ans. x' 2 x' 2 Q.11 Considering horizontal as X and vertical as Y and point of projection as origin ux = 20 uy = 0 ax = 0 ay = – g = – 10 vx = ux = 20 vy = – 10 t at t = 2 At the moment of collision with incline a angle with horizontal  = – vy tan  = x  – tan  = 10 (2) 20  tan  = 1   = 45° Displacement along horizontal x = uxt = 20 × 2 = 40 m Displacement along vertical y = 0 × 2 – 1 ×10(2)2 = – 20 m 2 Net displacement (OP) = = = 20 5m 1  1 2 1 1 5 Q.12 (a) h = 2 g 2  = 10  = 2 4 4 mts.   (b) An observed is train frame for coin ux = 0 uy = 0 ax = –0.8 ay = –10 t = 0.5 sec. y = –h 1 1  1 2 x = 0 × + (–0.8) +   = – 0.1 m 2 2  2  Coin lands 0.1m behind (c) An observed in ground frame ux = 2 ax = 0 t = 0.5 sec  1  x = ux t = 2 ×  2  = 1.0 m displacement of the coin is 1.0 m forward   Q.13 (a) x =10 t y = 10 t – 5 t2 dx dy vx = dt =10 vy = dt At t = 0 vx = 10 vy = 10 = 10 – 10 t v = = 10 m/s (b) a = dvx = 0 a = dvy = – 10 x dt y dt a = ay = –10 m/s2 (constant) (c) v v = s 10ˆi + 5ˆj = = 10ˆi + 5ˆj t 1 (av v ) = 5 m/s (d) Returning to same height means y = 0  10 t – 5t2 = 0  t = 2 m/s Q.14 (a) vy = uy – gt = u sin 60° – 10 t A the moment of collision t = 2 and vy = 0  0 = u 3 – 10 × 2  u = 2 (b) x = (u cos ) t 40 1 = 3 × 2 × 2 = 40/ m (c) Vertical displacement of ball I. y1 = uyt – 1 gt2 = 20×2 – 2 1 ×10 × (2)2 = 20 m (upward) 2 Vertical displacement of ball II y2 = – 1 gt2 = – 2 1 2 × 10 × (4) = – 20 m ( downward) height of tower h = (y1) + (y2) = 20 + 20 = 40 m Ans. Q.15 (a) (i) v0 1 t1 = 2 a t 2  a = 2v0 t1 2  40 = 20 = 4 m/s2 (ii) vm = a t1 = 4 × 20 = 80 m/s (b) a t2 = v0 v0  t2 =  40 = 4 = 10 c tan  = v0   = tan–1 (40)