Velocity of block B in the given arrangement is $300 m m / s e c$ towards right. Then velocity of A will be

200 m m / s e c

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100 m m / s e c

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450 m m / s e c

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150 m m / s e c

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Solution

The correct option is A $200 m m / s e c$

Assume rightward direction to be +ve
For pulley 1,
$v_{1} = v_{B} = \frac{v_{C} + v_{D}}{2}$
$v_{C} = 0$ since that end of the string is fixed.
$\Rightarrow v_{D} = 2 v_{B} - - (1)$

For pulley 2,
$v_{2} = v_{A} = \frac{v_{E} + v_{F}}{2}$
$v_{E} = v_{D}$
$\Rightarrow v_{A} = \frac{2 v_{B} + v_{F}}{2} - - (2)$ (using (1))

Pulley 3 is fixed
$\Rightarrow v_{G} = v_{H} = v_{A}$
But $v_{F} = v_{G}$
$∴ v_{F} = - v_{A} - - (3)$
(because direction is opposite to that of A)

Substituting (3) in (2),
$v_{A} = \frac{2 v_{B} - v_{A}}{2}$
$\Rightarrow 3 v_{A} = 2 v_{B}$
$\Rightarrow v_{A} = \frac{2 v_{B}}{3} = \frac{2 \times 300 m m / s e c}{3}$
$\Rightarrow v_{A} = 200 m m / s e c$

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