Question

Two pulley arrangements of figure given are identical. The mass of the rope is negligible. In fig (a), the mass $m$ is lifted by attaching a mass $2\text{m}$ to the other end of the rope. In fig (b), $m$ is lifted up by pulling the other end of the rope with a constant downward force $F=2mg$. The acceleration of $m$ in the two cases are respectively

• A. $3g,g$
• B. $\frac{g}{3},g$
• C. $\frac{g}{3},2g$
• D. $g,\frac{g}{3}$

Answer 1

The correct option is B $\frac{g}{3},g$

Let $a$ and $a^{\prime }$ be the accelerations in both cases respectively. Then for fig (a),
$T-mg=ma$ ... (i)
and $2mg-T=2ma$ ... (ii)
Adding (i) and (ii), we get
$mg=3ma$
$\therefore a=\frac{g}{3}$
For fig (b),
$T-mg=m{a}^{\prime }$ ... (iii)
and $2mg-T^{\prime }=0$ ... (iv)
Solving (iii) and (iv)
$a^{\prime }=g$
$\therefore a=\frac{g}{3}$ and $a^{\prime }=g$