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Variation of G Due to Height and Depth

In the system...

Question

In the system shown in figure $M_{1} > M_{2}$ and pulley and threads are ideal. System is held at rest by thread $B C$ . Just after thread $B C$ is burnt.

Acceleration

M_{1}

and

M_{2}

will be upward

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Magnitude of acceleration of both masses will be

\frac{M_{1} - M_{2}}{M_{1} + M_{2}} g

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Acceleration of

M_{1}

and

M_{2}

will be equal to zero

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Acceleration of

M_{1}

will be equal to zero, which that of

M_{2}

will be

\frac{M_{1} - M_{2}}{M_{2}} g

upward

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Solution

The correct option is D Acceleration of $M_{1}$ will be equal to zero, which that of $M_{2}$ will be $\frac{M_{1} - M_{2}}{M_{2}} g$ upward
As $M_{1} > M_{2}$ , so tension in thread connecting the block $B$ and support $C$ is $(M_{1} - M_{2}) g$ . When this thread is burnt, this tension disappears/ In this case tension in spring is $M_{1} g$ and it s elongated. So, tension in string connecting $A$ and $B$ is $M_{1} g$ . Hence, resultant force on $A$ remains zero because tension in string is balanced by tension in spring. The block $B$ has a net upward force $(M_{1} - M_{2}) g$ , so initial acceleration of block $B$ is
$\frac{(M_{1} - M_{2}) g}{M_{2}}$
Thus, initial acceleration of $M_{1}$ is zero and that of $M_{2}$ is $\frac{(M_{1} - M_{2}) g}{M_{2}}$ upward.

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