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Two blocks $A$ and $B$ each of same mass are attached by a thin inextensible string through an ideal pulley. Initially block $B$ is held in position as shown in figure. Now the block $B$ is released. Block $A$ will slide to right and hit the pulley in time $t_{A}$ . Block $B$ will swing and hit the surface in time $t_{B}$ . Assume the surface as frictionless

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Data are not sufficient to get relationship between $t_{A}$ and $t_{B}$ .

$t_{A} < t_{B}$

$t_{A} > t_{B}$

$t_{A} = t_{B}$

answer is B.

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Detailed Solution

In order to hit the surface block B has to travel ‘L’ distance, similarly to hit pulley block A must travel a distance of ‘L’

considering an arbitrary situation ,

$a_{A x} = \frac{T}{m} b u t a_{B x} = \frac{T \cos θ}{m} S o w e c a n s a y a_{A x} > a_{B x} a t a n y i n s \tan t i n t h e p r o c e s s .$

so $t_{A} < t_{B}$

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