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Question

The system shown in the figure is in equilibrium at rest and the spring and string are massless. Now the string is cut. The acceleration of masses 2m and m just after the string is cut will be:

A
3g2 upwards, g downwards
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B
g2 upwards, g downwards
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C
g upwards, 2g downwards
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D
2g upwards, g downwards
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Solution

The correct option is B g2 upwards, g downwards Initially, when the system is in equilibrium: FBD of system: By applying equilibrium condition on system in vertical direction ⇒FSpring=3mg Now, after the string is cut, T=0 i.e tension in string becomes zero instantaneously, while spring force will act at its initial value FSpring, due to inertia of spring. FBD of individual blocks after string is cut, represented below: Acceleration of block of mass m am=mgm ∴am=g (Downwards) Acceleration of block of mass 2m a2m=FSpring−2mg2m=mg2m ∴a2m=g2 (Upwards)

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