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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