Question

In the arrangement shown in the figure, the mass '$m$' is going upward with an acceleration of $1m/s^2$. The length of the rod is $100cm$. Pulley is frictionless and threads are massless . If the mass is set at the level of the lower end of the rod and released, how much time it will take for ball to reach the level of the upper end?

• A. $\sqrt{\frac{1}{3}}s$
• B. $\sqrt{\frac{2}{3}}s$
• C. $\frac{2}{3}s$
• D. $\frac{1}{3}s$

Answer 1

The correct option is B $\sqrt{\frac{2}{3}}s$

Let acceleration of the mass be $a_1$ and for the rod be $a_2$
Given that $a_1=1m/s^2$

Since pulley 1 is fixed,
$a_A=a_B=a_1$

Considering pulley 2,
$\frac{a_C+a_D}{2}=a_B=a_1$
Since D end is fixed, $a_D=0$ and $a_C=a_2$
$\Rightarrow a_1=\frac{a_2+0}{2}=\frac{a_2}{2}$
$\Rightarrow a_2=2a_1=2m/s^2$

Now for the rod w.r.t the mass $m$, we know
Relative initial velocity $u_{rel}=0$
Relative acceleration $a_{rel}=a_2-a_1$
$=2-(-1)=3m/s^2$ ($\because$ in opposite directions)
Relative displacement $S_{rel}=100cm=1m$ (length of the rod)

Using $2^{nd}$ equation of motion,
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow 1=\frac{1}{2}\times \left(3\right)\times t^2$
$\Rightarrow t=\sqrt{\frac{2}{3}}s$