System shown in figure is in equilibrium, find the magnitude of net change in the string tension between two masses just after, when one of the springs in cut, Mass of both the blocks is same and equal to m spring constant of both springs is k.
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Options

A. `(mg)/(2)`
B. `(mg)/(4)`
C. `(mg)/(3)`
D. `(3mg)/(2)`

Answer

A

Explanation

`T_(i)=gh`......(i)
`2kx =2mg`
`:. kx =mg`
One `kx` force (acting is upward directon) is suddenly removed. So, net downward force on system will be `kx` or `mg`. Threrfore, net downward acceleration of system,
`a=(mg)/(2m) =g/2`
Free body diagram of lower block gives the equation,
`mg-T_(f) =ma=(mg)/(2)`
`:.T_(f)=(mg)/(2)`
From these two equation, we get
`DeltaT =(mg)/(2)`.