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(C) For a system of two masses $m_1$ and $m_2$ connected by an inextensible string over a smooth pulley, the acceleration $a$ is given by the formula:

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$\frac{n+1}{n-1}$

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(C) For a system of two masses $m_1$ and $m_2$ connected by an inextensible string over a smooth pulley, the acceleration $a$ is given by the formula:$a = \frac{m_2 - m_1}{m_2 + m_1} g$Given that the acceleration $a = \frac{g}{n}$, we equate the two expressions:$\frac{g}{n} = \frac{m_2 - m_1}{m_2 + m_1} g$$\frac{1}{n} = \frac{m_2 - m_1}{m_2 + m_1}$Cross-multiplying gives:$m_2 + m_1 = n(m_2 - m_1)$$m_2 + m_1 = n m_2 - n m_1$Rearranging the terms to group $m_1$ and $m_2$:$m_1 + n m_1 = n m_2 - m_2$$m_1(n + 1) = m_2(n - 1)$Therefore, the ratio $\frac{m_2}{m_1}$ is:$\frac{m_2}{m_1} = \frac{n + 1}{n - 1}$

An inextensible string passing over a smooth pulley connects two blocks of masses $m_1$ and $m_2$ $(m_2 > m_1)$ vertically. If the acceleration of the system is $(\frac{g}{n})$, then the ratio of masses $(\frac{m_2}{m_1})$ is

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