Question

The pulley arrangements shown in figure are identical. The mass of the rope being negligible. In case-$\text{I}$, the mass $m$ is lifted by attaching a mass $2m$ to the other end of the rope. In case-$\text{II}$, the mass $m$ is lifted by pulling the other end of the rope with constant downward force $F=2mg$, where $g$ is acceleration due to gravity. The acceleration of mass $m$ in case-$\text{I}$ is

• A. more than that in case II
• B. equal to that in case II
• C. Zero
• D. less than that in case II

Answer 1

The correct option is D less than that in case II

Let the accelerations in both cases are $a$ and $a\text{'}$ respectively.
Case $\left(i\right):$
Tension is given as,
$T-mg=ma$ .....$\left(1\right)$
$2mg-T=2ma$ .....$\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$
$mg=3ma$
$\Rightarrow a=g/3$

Case $\left(ii\right):$
Tension is given as,
Here, $F=T$
$T^{\prime }-mg=m{a}^{\prime }$ .....$\left(3\right)$
$T^{\prime }=2mg$ .....$\left(4\right)$
On solving $\left(3\right)$ and $\left(4\right)$
$a^{\prime }=g$
On comparing a of case of $\left(i\right)$ case of $\left(ii\right)$,
$\frac{a}{a^{\prime }}=\frac{1}{3}$ or $a=\frac{1}{3}{a}^{\prime }$
Hence , option $\left(C\right)$ is correct.