Question

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

• A. $1.4s$
• B. $2.45s$
• C. $3.25s$
• D. $5s$

Answer 1

Answer: (A)

Let $a_1$ and $a_2$ be accelerations of a ball (upward) and rod (downward), respectively.

Clearly, from the diagram
$2a_1-a_2$ ...(i)
Now, for the ball
$2T-\frac{9}{5}mg-\frac{9}{5}ma$ ...(ii)
and for the rod, $mg-T=m{a}_2$ ...(iii)
On solving equations (i) and (iii), we get
$a_1=\frac{g}{29}m{s}^{-2}\uparrow$ (upward)
$a_2=\frac{2g}{29}m{s}^{-2}\downarrow$ (downward)
So, acceleration of ball w.r.t rod $=a_1+a_2=\frac{3g}{29}$
Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is $1m$ .
Using the equation of motion,
$s=ut+\frac{1}{2}a{t}^2$
$s=0+\frac{1}{2}\times \frac{3\times 10}{29}{t}^2$
$t=\sqrt{\frac{58}{30}}=1.4s\left(approx\right)$