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A block tied between two identical sprin...

A block tied between two identical springs is in equilibrium. If upper spring is cut then the acceleration of the block just after cut is `6m//s^(2)` downwards. Now, if instead of upper spring, lower spring is being cut then the acceleration of the block just after the cut will be.

A

`4 m//s^(2)` downwards

B

`6 m//s^(2)` downwards

C

`4 m//s^(2)` upwards

D

`6 m//s^(2)` upwards

Text Solution

Verified by Experts

The correct Answer is:
A

`F_(1)+F_(2)=mg` ...(i)
If upper spring is cut
`a=(mg-F_(1))/(m)=6rArrF_(1)=4m`
If lower spring is cut `therefore F_(2)=6m`
`a=(mg-F_(2))/(m)=(10m-6m)/(m)=4`
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