11.PRACTICE TEST-3 (Solution-Paper-2)

As the linear momentum of any system = MVCM  The linear momentum of the train = 2MV  d Ans. y 5. dt = 2   = 2t Let BP = a  x = OM = a sin = a sin(2t) Hence M executes SHM within the given time period and its acceleration is opposite to ‘x’ that means towards left x I1  a2  2 (3)2 (8)2 6. I2 a2  2 = (2)2 (12)2 = 1 2 2 → → 1 → → → 7. → = 0 q v  r and → = qr  → =   →  → v  E = B 4 r 3 d E 4 0 r 3 B 0 0 (v E) c2 8. dq = R 2i 20 a 𝑙  f = – i  2a  0ia.𝑙n2 = R = R.2 n  a R .   9. Initially 𝑙 H = 3mg × 2 2 2m𝑙2 m𝑙2  3𝑙   = + + m   H 3 12   3 = 2 m𝑙2 = H    = g 𝑙 Hing Force 3 N1 = 3mg cos 60° = 2 mg 3mgsin60° – N2 Hinge force = g = 3m × 𝑙 × = 𝑙  N = mg 3 mg 2 10. → ˆ ˆ V  yi  xj dx dy dy x  dt = y & dt = – x  dx = – y = x2 + y2 = c (circle) speed = = = constant → = dy ˆ– dx ˆ a dt i dt j → = – x ˆ– y ˆ  → ˆ ˆ (conservative) a i j F   mxi  myj 15. mgh – FRd = 0 mgh d = FR = 20 m d 10 = 2 m Ans. r 2.0 16. 400 = d 6r 2 0  3r 2 0 .....(i)  9  C = 4 d 4 d 4 400 = 900 F C = 100 × 9 F Hence X = 9 17. x 𝑙 / 2 q  2 q 1  𝑙 3 q 2 M =  2. dx. .x =   = 𝑙 𝑙 2 0 𝑙 3  2  24 18. N1 N1 sin 30° N2 cos 60° N1 cos 30° N2 30° 60° 2 1 N2 sin 60° 60° 30° N1 sin 30° = N2 sin 60° N1 cos 30° + N2 cos 60° = mg Solving above equation N = mg  10  10  50 2 2 2 19. In case A and B acceleration is constant but speed first decreases and then increases. In case C and D, the velocity does not change sign hence direction of acceleration is constant. Speed and magnitude of acceleration decreases with time.