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Two blocks $A$ and $B$ of equal mass are connected by a light inextensible taut string passing over two light smooth pulleys fixed to the blocks. The parts of the string not in contact with the pulleys are horizontal. A horizontal force $F$ is applied to the block $A$ as shown. There is no friction, then

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The acceleration of $A$ will be more that of $B$

The sum of rate of changes of momentum of $A$ and $B$ is greater than the magnitude of $F$ .

The sum of rate of changes of momentum of $A$ and $B$ is equal to the magnitude of $F$ .

The acceleration of $A$ will be less than that of $B$

answer is A, C.

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Detailed Solution

$2 T \cdot X_{A} = 3 T \cdot X_{B} X_{A} = \frac{3}{2} X_{B} {a_{A}}_{} = 1.5 a_{B}$

Net external force on the system is $F + T (> F)$

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