Three small discs are connected with two identical massless rods as shown in fig. the rods are pinned to the discs such that angle between them can change freely. The system is placed on a smooth horizontal surface with discs A and B touching a smooth wall and the angle ACB being $90^{o}$ . A force F is applied to the disc C in a direction perpendicular to the wall. Find acceleration of disc B immediately after the force starts to act. Masses of discs are $m_{A} = m; m_{B} = 2 m; m_{C} = m$
[Wall is perpendicular to the plane of the fig.]

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$\frac{F}{7 m}$

$\frac{F}{5 m}$

$\frac{F}{3 m}$

$\frac{F}{m}$

answer is C.

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Detailed Solution

Component of acceleration of C can be assumed along the two roads ( the rods are……)as $a_{1} a n d a_{2}$ since rods are rigid
$a_{1} = \frac{a_{A}}{\sqrt{2}}$
$a_{2} = \frac{a_{B}}{\sqrt{2}}$
For $A \frac{T_{A}}{\sqrt{2}} = m a_{A}$ ……….. (1)
For $B \frac{T_{B}}{\sqrt{2}} = m a_{B}$ ………. (2)
For $C \frac{F}{\sqrt{2}} - T_{A} = m a_{1}$ ………(3)
And $\frac{F}{\sqrt{2}} - T_{B} = m a_{2}$ ………(4)
Using (2) and (4)
$\frac{F}{\sqrt{2}} - 2 \sqrt{2} m a_{B} = m a_{2}$
$\frac{F}{\sqrt{2}} - 2 \sqrt{2} m a_{B} = m = \frac{a_{B}}{\sqrt{2}} (∵ a_{2} = \frac{a_{B}}{\sqrt{2}})$
$∴ \frac{F}{\sqrt{2}} = m a_{B} (\frac{1}{\sqrt{2}} + 2 \sqrt{2})$
$\frac{F}{5 m} = a_{B}$