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The pulley arrangements of figures(a) and (b) are identical .The mass of the rope is negligible .In figure(a) ,the mass m is lifted up by attaching a mass 2m to the other end of the rope .In figure(b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2mg . Calculate the accelerations in the two cases.

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$a = \frac{g}{3}, a^{1} = g$

$a = \frac{2 g}{3}, a^{1} = 4 g$

$a = \frac{g}{3}, a^{1} = 3 g$

$a = \frac{4 g}{3}, a^{1} = 2 g$

answer is A.

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Detailed Solution

$\begin{array}{lc} 2 g - T = 2 a & 2 mg - T = 0 \\ \frac{T - g = a}{g = 3 a} & \frac{T - mg = ma}{mg = ma} \\ a = \frac{g}{3} & a^{1} = g \end{array}$

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