In figure shown, pulley are ideal $m_{1} > 2 m_{2}$ . Initially the system is in equilibrium and string connecting $m_{2}$ to rigid support below is cut. Find the initial acceleration of $m_{2}$ ?

(\frac{m_{1} - 2 m_{2}}{2 m_{2}}) g

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(\frac{m_{1} - m_{2}}{2 m_{2}}) g

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(\frac{m_{2} - m_{1}}{2 m_{2}}) g

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(\frac{m_{1} + 2 m_{2}}{2 m_{2}}) g

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Solution

The correct option is B $(\frac{m_{1} - 2 m_{2}}{2 m_{2}}) g$
In equilibrium:
On the left pulley:
Force in the spring $=$ tension in the string $= \frac{m_{1} g}{2}$

On the right pulley:
Tension in the string will same as on the right pulley.
$T = \frac{1}{2} m_{1} g$

Tension in the string below $m_{2}$
$T^{'} = T - m_{2} g$
$⟹ T^{'} = \frac{1}{2} m_{1} g - m_{2} g$

As the string below $m_{2}$ is cut:
Additional force on $m_{2}$ will be equal to tension $T^{'}$ on the string below $m_{2}$ .

$⟹ m_{2} a = T^{'}$ ; where a is the acceleration of $m_{2}$
$⟹ m_{2} a = \frac{1}{2} m_{1} g - m_{2} g$

$⟹ a = (\frac{m_{1} - 2 m_{2}}{2 m_{2}}) g$

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In figure shown, pulley are ideal m1>2m2. Initially the system is in equilibrium and string connecting m2 to rigid support below is cut. Find the initial acceleration of m2?

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