In the arrangement shown in figure, pulleys are massless and frictionless and threads are inextensible, block of mass $m_{1}$ will remain at rest, if:

\frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}}

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m_{1} = m_{2} = m_{3}

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\frac{1}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}}

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\frac{1}{m_{1}} = \frac{1}{m_{2}} + \frac{3}{m_{3}}

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Solution

The correct option is A $\frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}}$
$\begin{matrix} a = [\frac{m_{3} - m_{2}}{m_{2} + m_{3}}] g & m_{3} > m_{2} T = \frac{2 m_{2} m_{3} g}{m_{2} + m_{3}} T^{'} = 2 T = \frac{4 m_{2} m_{3} g}{m_{2} + m_{3}} m_{1} g = \frac{4 m_{2} m_{3} g}{m_{2} + m_{3}} \frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}} \end{matrix}$

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