A frictionless cart of mass $M$ carries two other frictionless carts having masses $m_{1}$ and $m_{2}$ connected by a string passing over a pulley as shown in figure. The horizontal force that must be applied on $M,$ so that $m_{1}$ and $m_{2}$ do not move relative to it will be

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

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

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(M + m_{1}) [\frac{(m_{1} + m_{2})}{m_{2}}] g

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(M + m_{2}) [\frac{m_{2}}{(m_{1} + m_{2})}] g

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Solution

The correct option is A $(M + m_{1} + m_{2}) (\frac{m_{2}}{m_{1}}) g$

$F = (M + m_{1} + m_{2}) a$
$a = \frac{F}{(M + m_{1} + m_{2})}$
pseudo force on $m_{1},$
$F^{'} = m_{1} a = \frac{m_{1} F}{(M + m_{1} + m_{2})} = T . . . . (i)$

For $m_{2}, T = m_{2} g . . . . (i i)$
from $(i) & (i i)$
$F^{'} = m_{2} g$
$\frac{m_{1} F}{(M + m_{1} + m_{2})} = m_{2} g$
$∵ F = \frac{m_{2} g}{m_{1}} (M + m_{1} + m_{2})$
$∵ F = \frac{m_{2} g}{m_{1}} (M + m_{1} + m_{2})$
So option (A) is correct.

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