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System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass $2 m$ and $m$ just after the string is cut will be
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Verified Answer
The correct answer is:
$g / 2$ upwards, $g$ downwards
$g / 2$ upwards, $g$ downwards
Initially under equilibrium of mass ' $m$ '
$$
T=m g
$$
Now, the string is cut. Therefore, $T=m g$ force is decreased on mass $m$ upwards and downwards on mass $2 m$.
$$
a_m=\frac{m g}{m}=g \text { (downwards) and }
$$
$$
a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} \text { (upwards) }
$$
$$
T=m g
$$
Now, the string is cut. Therefore, $T=m g$ force is decreased on mass $m$ upwards and downwards on mass $2 m$.
$$
a_m=\frac{m g}{m}=g \text { (downwards) and }
$$
$$
a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} \text { (upwards) }
$$
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