The two pulley arrangements shown in the figure 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 of 2mg. The ratio of accelerations in two cases will be

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1:4

1:3

1:2

1:1

answer is C.

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Detailed Solution

For first case,

$a_{1} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g = \frac{2 m - m}{m + 2 m} = \frac{g}{3} . . . . . . . (i)$

For second case,
From free body diagram of m,

$\begin{array}{l} m a_{2} = T - m g \\ m a_{2} = 2 m g - m g [As T = 2 m g] \\ a_{2} = g \end{array}$

$From (i) and (ii), \frac{a_{1}}{a_{2}} = \frac{g / 3}{g} = 1 / 3$

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