In the figure, mass of a ball is $\frac{9}{5}$ times mass of the rod. Length of rod is 1 m. The level of ball is same as rod level. Find out time taken by the ball to reach at upper end of rod.                               [AIIMS 2018]

detailed solution

Correct option is A

Let a1 and a2 be accelerations of a ball (upward) and rod (downward), respectivelyClearly, from the diagram,2a1 = a2         ......(i)Now, for the ball,2T - 95 mg = 95 ma1         ......(ii)and for the rod, mg - T = ma2    …………..(iii)On solving Eqs. (i), (ii) and (iii), we geta1=g29 m/s2↑( upward )a2=2g29 m/s2↓ (downward) So, acceleration of ball w.r.t. rod = =a1+a2=(3g/29)m/s2Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is 1m.Using second equation of motion,   s=ut+12at21=0+12×3×1029t2⇒  t=58/30=1.4 s (approx)