Question

In the pulley system, the mass of ball is $1.2$ times greater than the mass of the rod.The length of the rod is $50\, cm$. The ball is set on the same level as the lower end ofthe rod and then released. What is the acceleration of the rod with which it comesdown?
Assume the pulleys and threads are massless and friction force is neglected. (Use $g =10\, m / s 2$ )

• A. $4\, m / s$
• B. $3 \,m / s$
• C. $2\, m / s$
• D. $5\, m / s$

Answer 1

Answer: (D)

The free body diagram for given system is as shown below

$\Rightarrow m a_{1}=2 T-m g \ldots$ (i)

$\Rightarrow M a_{2}=M g-T \ldots (ii) $
As, thread is uniform and of constant length,
so $2 a_{1}=a_{2}$
From Eq. (i), we get
$\frac{m a_{2}}{2}=2 T-m g \ldots (ii) $
Multiplying Eq. (ii) by $2$ and adding to Eq. (iii), we get
$2 M a_{2}+\frac{m a_{2}}{2}=2 M g-m g $
$\Rightarrow a_{2}=\frac{(2 M-m)}{\left(2 M+\frac{m}{2}\right)} g=\frac{\left(2-\frac{m}{M}\right)}{\left(2+\frac{m}{2 M}\right)} g$
Here, $\frac{m}{M}=1.2$ and $g=10 ms ^{-2}$
$\therefore a_{2}=\frac{(2-1.2)}{\left(2+\frac{1.2}{2}\right)} \times 10 $
$=\frac{0.8}{26} \times 10=\frac{40}{1.3} \propto 3\, m / s ^{2}$