Water is filled to height `h` in a fixed vertical cylinder placed on horizontal surface. At time `t = 0` a small hole a drilled at a height `h//4` from bottom of cylinder as shown. The cross section area of hole is a and the cross-section area of cylinder is `A` such that `A gtgt a`.

The duration of time for which water flows out of hole is.

Options

A. `(A)/(a)sqrt((3h)/(2g))`
B. `(a)/(A)sqrt((3h)/(2g))`
C. `(A)/(a)sqrt((2h)/(3g))`
D. None of these

Answer

A

Explanation

The initial velocity of water coming out of hole is horizontal and hole is at height `(h)/(4)` from ground. Hence time taken by water to reach ground is `t = sqrt((2(h//4))/(g)` which remains constant.

`:. x = vt`
where `v` is velocity of eff ux...
Since `v` decreases with time `x` will decrease.
Let `y` be the height of water surface above hole
`:. -(dy)/(dt) = (av)/(A) = (asqrt(2gy))/(A)`
`:. overset(0)underset(3h//4)(int)(dy)/(sqrt(2gy)) = overset(t)underset(0)(int)-(a)/(A)dt`
`:. t = (A)/(a)sqrt((3h)/(2g))`