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Standard XIII

Physics

Pulley

Immediately o...

Question

Immediately on cutting the taut string, the acceleration of one of the blocks is found to be $2 m s^{- 2} .$ If $m_{1} = 6 k g, m_{2}$ is equal to $(g = 10 m s^{- 2})$

4 k g

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5 k g

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2 k g

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8 k g

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Solution

The correct option is B $5 k g$
At rest,
$T^{'} = (m_{1} - m_{2}) g$
$∴ m_{1} > m_{2}$
On cutting $T^{'} = 0$ ,

hence net force on $m_{2}$ is $(m_{1} - m_{2}) g$
$m_{2} a = (m_{1} - m_{2}) g$
$∴ a = (\frac{m_{1} - m_{2}}{m_{2}}) g$ (upwards)
After cutting the string spring force doesn't change , so $m_{1}$ 's acceleration is zero.
$∴ (\frac{m_{1} - m_{2}}{m_{2}}) g = 2$ $(∵ a = 2)$
$\Rightarrow \frac{m_{1}}{m_{2}} = 1.2$
$\Rightarrow \frac{6}{m_{2}} = 1.2$
$m_{2} = 5 k g$

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Immediately on cutting the taut string, the acceleration of one of the blocks is found to be 2 ms−2. If m1=6 kg,m2 is equal to (g=10 ms−2)

Immediately on cutting the taut string, the acceleration of one of the blocks is found to be $2 m s^{- 2} .$ If $m_{1} = 6 k g, m_{2}$ is equal to $(g = 10 m s^{- 2})$