The acceleration of the blocks A and B in the situations shown in the figure

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$\frac{3}{7} g downwards, \frac{2}{7} g upwards$

$\frac{2 g}{7} downwards, \frac{2 g}{7} upwards$

$\frac{g}{7} downwards, \frac{2}{7} g upwards$

$\frac{2}{7} g downwards, \frac{g}{7} upwards$

answer is A.

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Detailed Solution

$2 T - 5 g = 5 a \to (1) 4 g - T = 8 a \to (2) F r o m e q u a t i o n (1) & (2) a = \frac{3}{21} g = \frac{g}{7} ∴ a_{A} = \frac{g}{7} ↑, a_{B} = 2 a = \frac{2g}{7} ↓$

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