Two particles having mass ratio n : 1 are interconnected by a light in extensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is

(n - 1)^{2} g

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{(\frac{n + 1}{n - 1})}^{2} g

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{(\frac{n - 1}{n + 1})}^{2} g

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(\frac{n + 1}{n - 1}) 9

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Solution

The correct option is C ${(\frac{n - 1}{n + 1})}^{2} g$
Given $,$
$\frac{m_{1}}{m_{2}} = \frac{n}{1} = n$
Each mass have the acceleration $a = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}}$
however $m_{1}$ which is heavier will have the will have acceleration $a_{1}$ vertically down while the lighter mass $m_{2}$ will have acceleration $a_{2}$ vertically up $\to a_{2} = - a_{1}$
The acceleration or the centre of mass of the system $,$ $a_{c m} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$
given that $a_{2} = - a_{1} \to a_{c m} = \frac{(m_{1} - m_{2}) a_{1}}{m_{1} + m_{2}} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \times \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}} = \frac{{(m_{1} - m_{2})}^{2} g}{m_{1} + m_{2}}$
Since $\frac{m_{1}}{m_{2}} = n$ diving by $m_{2}$ and simplifying
$\Rightarrow a_{c m} = {(\frac{n - 1}{n + 1})}^{2} g$
Hence,
option $(C)$ is correct answer.

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