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Pulley

Two weight ...

Question

Two weight $W_{1} a n d W_{2}$ are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with an acceleration g/2. The tension in the string will be:-

A

\frac{4 W_{1} W_{2}}{W_{1} + W_{2}}

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B

\frac{W_{1} W_{2}}{W_{1} + W_{2}}

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C

\frac{W_{1} W_{2}}{3 (W_{1} + W_{2})}

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D

none of these

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Solution

The correct option is D $\frac{4 W_{1} W_{2}}{W_{1} + W_{2}}$
$\begin{matrix} L e t m_{1} = \frac{w_{1}}{g} [w = m g] m_{2} = \frac{w_{2}}{g} . . . . . . . . . . . . (1) m_{2} g - T = m_{2} (a - g) T - m_{1} g = m_{1} (a + g) . . . . . . . . . (2) m_{2} g - m_{1} g = m_{2} a - m_{2} g + m_{1} a + m_{1} g 2 m_{2} g = 2 m_{1} g + a (m_{1} + m_{2}) a = \frac{2 g (m_{2} - m_{1})}{(m_{2} + m_{1})} . . . . . . . . . . (3) b y e q . (1), (2) & (3), T = m_{1} g + m_{1} a + m_{1} g = m_{1} (2 g + a) = \frac{w_{1}}{g} [2 g + 2 g (\frac{m_{2} - m_{1}}{m_{2} - m_{1}})] = w_{1} [\frac{2 (m_{2} + m_{1}) 2 (m_{2} - m_{1})}{(m_{2} + m_{1})}] = w_{1} ⎡ ⎢ ⎢ ⎢ ⎣ \frac{4 \frac{w_{2}}{g}}{\frac{w_{2} + w_{1}}{g}} ⎤ ⎥ ⎥ ⎥ ⎦ = \frac{4 w_{1} + w_{2}}{w_{1} + w_{2}} \end{matrix}$

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