Velocity of water flow in downward direction relative to cylindrical tank is `v_(e)`. Acceleration at any time t will be: (where `mu` is rate of change in mass at any time t)

Answer

A

Explanation

For block `Mg-T=M(dv)/(dt) ..........(1)`
For cylindrical tank, `T=muv_(g)-(M_(0)-mut)g=(M_(0)-mut) (dv)/(dt) ......(2)`
From (1) and (2), we get
`Mg+muv_(e)-(M-mut)g=(2M-mut) (dv)/(dt ........(3)`
`(muv_(e)+mugt)=(2M-mut) (dv)/(dt) ........(4)`
`(dv)/(dt)=(mu(v_(e)+gt))/((2M-mut)) " "therefore (dv)/(dt)=(mu(v_(e)+gt))/((2M-mut))`