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A constant force $F = 2 m g$ is applied on the block of mass $m_{2}$ as shown in figure. The string and the pulley are light and the surface of the table is smooth. given $m_{2} = 2 m$ , $m_{1} = m$
Column-I Column-II
A Acceleration of block $m_{2}$ P $\frac{4 m g}{3} \sqrt{2}$
B Tension in the string attached to blocks Q $\frac{4 m g}{3}$
C Acceleration of block $m_{1}$ R $\frac{g}{3}$ up
D Pulling force on pulley due to string S $\frac{g}{3}$ left

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$A - P; B - Q; C - R; D - S$

$A - S; B - Q; C - R; D - P$

$A - Q; B - R; C - S; D - P$

$A - Q; B - S; C - P; D - R$

answer is B.

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Detailed Solution

mg-T=ma..(1)

T=2ma...(2)

T= $\frac{4 mg}{3}$

Acceleration of block $m_{2}$ = $\frac{g}{3}$ left
Acceleration of block $m_{1}$ = $\frac{g}{3}$ up

Pulling force on pulley due to string will be, T left and T down.

Resultant = $\frac{4 m g}{3} \sqrt{2}$

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