Two blocks of masses m and M $=$ 2m are connected by means of a metal wire of cross-sectional area A passing over a frictionless fixed pulley as shown in figure. The system is then released.
The common acceleration of the blocks is

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\frac{g}{3}

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\frac{2 g}{3}

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\frac{3 g}{2}

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Solution

The correct option is B $\frac{g}{3}$
As the pulley is frictionless and fixed so the tension on the string on both of the pulley will be equal i.e T (say) and also the blocks have same acceleration, a.
From FBD for m : $m a = T - m g . . . (1)$
and for M : $M a = M g - T . . . . (2)$
Adding two equations, $(m + M) a = (M - m) g$
or $a = \frac{M - m}{M + m} g = \frac{2 m - m}{2 m + m} g = g / 3$ as $M = 2 m$

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