Two bodies with masses $m_{1}$ and $m_{2}$ $(m_{1} > m_{2})$ are joined by a string passing over a fixed pulley. The centres of gravity of the two bodies are initially at the same height. Assume mass of the pulley and weight of the thread negligible. Then the downwards acceleration of $m_{1}$ , is

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

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

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

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

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Solution

The correct option is A $(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g$
Here, we are given that the mass $m_{1}$ and $m_{2}$ pass over the fixed pulley. So the forces corresponding to the mass $m_{1}$ and $m_{2}$ are,
$m_{1} g - T = m_{1} a . . . . . (1)$
$T - m_{2} g = m_{2} a . . . . . (2)$

Adding equations (1) and (2), we get
$(m_{1} - m_{2}) g = (m_{1} + m_{2}) a$
$\Rightarrow a = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g$
The downward acceleration $= (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g$

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