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Com of Solid Bodies

In the arrang...

Question

In the arrangement, shown in figure, pulleys are massless and frictionless and threads are inextensible block of mass $m_{2}$ will remain at rest if.

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

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

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Solution

The correct option is B $\frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}}$
Let, the tension in string connected to $m_{1}$ is $T$ and $m_{2}$ and $m_{3}$ is $T_{1}$ .

$T = m_{1} g$
$2 T_{1} = T$
$T_{1} = \frac{T}{2}$

let, the acceleration of the mass $m_{2}$ be
$m_{2} g - T_{1} = m_{2} a$ ......(I)
$T_{1} - m_{3} g = m_{3} a$ ......(II)

After solving,

$m_{2} g - m_{3} g = (m_{2} + m_{3}) a$
$g (m_{2} - m_{3}) = (m_{2} + m_{3}) a$

Now,

$a = \frac{g (m_{2} - m_{3})}{m_{2} + m_{3}}$

Now, put the value of $a$ in equation (II)

$T_{1} - m_{3} g = m_{3} (\frac{g (m_{2} - m_{3})}{m_{2} + m_{3}})$

$T_{1} = m_{3} \frac{g (m_{2} - m_{3})}{m_{2} + m_{3}} + m_{3} g$

$T_{1} = m_{3} g (\frac{2 m_{2}}{m_{2} + m_{3}})$

$T_{1} = \frac{2 m_{2} m_{3} g}{m_{2} + m_{3}}$

Now, put the value of $T_{1}$

$\frac{T}{2} = \frac{2 m_{2} m_{3} g}{m_{2} + m_{3}}$

$T = \frac{4 m_{2} m_{3} g}{m_{2} + m_{3}}$

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

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

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

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

Therefore the block of mass $m_{2}$ at rest if $\frac{4}{m_{1}} = \frac{1}{m_{3}} + \frac{1}{m_{2}}$

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In the arrangement, shown in figure, pulleys are massless and frictionless and threads are inextensible block of mass m2 will remain at rest if.

In the arrangement, shown in figure, pulleys are massless and frictionless and threads are inextensible block of mass $m_{2}$ will remain at rest if.