System shown in the 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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System shown in the figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass 2m and m just after the string is cut will be

MARKS Content Team · Accepted Answer

Correct answer: (A) g/2 upwards, g downwards. Initially, both the blocks were under equilibrium, so for mass m, T=mg, and for mass 2m, F_s=T+2mg=3mg Now, after the string is cut, the tension in it becomes zero. Therefore, for m, mg-0=ma ⇒a=g(downward) and for 2m, 2mg+0-F_s=2mb2mg+0-3mg=2mb⇒b=-g/2 i.e., g/2 upward.

System shown in the 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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