Two blocks $A$ and $B$ of the same mass are connected through a string and arranged as shown in the figure. When the system is released from rest and there is no friction, then tension in the string is $\frac{m g}{p}$ . Then $p =$

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Solution

The correct option is C 4
Pulling force acting on $B = m g$ and that on $A = m g sin θ$
Since $m g > m g sin θ$ , block $B$ will move down.

For $B$ :
$m g - T = m a$
For $A$ :
$m g sin θ + T = m a$
Hence $m g sin θ + T = m g - T$
$\Rightarrow T = \frac{m g}{2} (1 - sin θ) = \frac{m g}{4}$

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