Question

The pulley arrangements of the figure. (a) and (b) are identical. The mass of the rope is negligible. In (a) the mass $m$ is lifted up by attaching a mass $2m$ to the other end of the rope. In (b), $m$ is lifted up by pulling the other end of the rope with a constant downward force $F = 2 \, m g$ . The acceleration of $m$ in both cases is?

• A. $g/3$ , $g/3$
• B. $g/3$ , $g$
• C. $g/4$ , $g/3$
• D. $g/4$ , $g$

Answer 1

Answer: (B)

Case A
For mass $m$
$T - m g = m a$ (i)
For mass $2m$
$2 \, m g - T = 2 \, m a$ (ii)

From (i) and (ii), we get –
$a = \frac{g}{3}$ …(iii)
Case B
$T-mg=ma'$
$\Rightarrow \, 2 m g - m g = m a '$ $\left[\right. \because \, \, T = 2 m g \left]\right.$
$\therefore \, \, a ' \, = g$ …(iv)