A light string passes over a frictionless pulley. To one of its ends a mass of $8 k g$ is attached. To its other end two masses of $7 k g$ each are attached. The acceleration of the system will be

10.2 g

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5.10 g

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20.36 g

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0.27 g

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Solution

The correct option is D $0.27 g$
From free body diagram of body $A$ ,
$m_{1} a = T_{1} - m_{1} g$ ....(i)
For body $B$ ,
$m_{2} a = m_{2} g + T_{2} - T_{1}$ ....(ii)
For body $C$ ,
$m_{3} a = m_{3} g - T_{2}$ .....(iii)
On solving equations (i), (ii) and (iii), we get
$a = \frac{[(m_{2} + m_{3}) - m_{1}] g}{m_{1} + m_{2} + m_{3}}$
As, $m_{1} = 8 k g$ , $m_{2} = m_{3} = 7 k g$
$a = \frac{6}{22} g = 0.27 g$

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A light string passes over a frictionless pulley. To one of its ends a mass of 8kg is attached. To its other end two masses of 7kg each are attached. The acceleration of the system will be

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A light string passes over a frictionless pulley. To one of its ends a mass of $8 k g$ is attached. To its other end two masses of $7 k g$ each are attached. The acceleration of the system will be