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]

Let a₁ and a₂ be accelerations of a ball (upward) and rod (downward), respectively

Clearly, from the diagram,

$2a_1 = a_2 ......\left(\mathrm{i}\right)$

Now, for the ball,

$2T - \frac{9}{5} mg = \frac{9}{5} m{a}_1 ......\left(\mathrm{ii}\right)$

and for the rod, mg - T = ma₂    …………..(iii)

On solving Eqs. (i), (ii) and (iii), we get

$a_1=\frac{g}{29} \mathrm{m}/{\mathrm{s}}^2\uparrow \left(\text{ upward }\right)$

$a_2=\frac{2g}{29} \mathrm{m}/{\mathrm{s}}^2\downarrow \text{ (downward) }$

So, acceleration of ball w.r.t. rod = $=a_1+a_2=(3g/29)\mathrm{m}/{\mathrm{s}}^2$

Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is 1m.

Using second equation of motion,   $s=ut+\frac{1}{2}a{t}^2$

$1=0+\frac{1}{2}\times \frac{3\times 10}{29}{t}^2$

$\Rightarrow t=\sqrt{58/30}=1.4 \mathrm{s}\text{ (approx) }$