The acceleration of the block of mass M in the situation is as shown in the figure. All the
surfaces are frictionless and the pulleys and the string are light

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$\frac{2 g}{3} u p t h e p l a n e$

$\frac{3 g}{2} u p t h e p l a n e$

$\frac{g}{3} d o w n t h e p l a n e$

$\frac{g}{3} u p t h e p l a n e$

answer is D.

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Detailed Solution

$T - M g \sin 30 ° = M a \to (1) 2 M g - 2 T = 2 M \times \frac{a}{2} \to (2) E q (1) \times (2), a n d f r o m (1) & (2) 2 T - 2 M g \sin 30 = 2 M a 2 M g - 2 T = M a 2 M g - M g = 3 M a \Rightarrow a_{M} = \frac{g}{3} ↑$

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