A vessel of area of cross-section A has liquid to a height H . There is a hole at the bottom of vessel having area of cross-section a . The time taken to decrease the level from `H_(1)` to `H_(2)` will sec

Options

A. `(2A)/( pi a ^(2)) sqrt ((h)/(g ))`
B. `(sqrt2 A )/( pi a ^(2)) sqrt ((h )/(g))`
C. ` (2 sqrt2 A )/( pi a ^(2)) sqrt((h )/(g))`
D. `(A)/( sqrt2 pi a ^(2)) sqrt ((h)/(g))`

Answer

B

Explanation

Let the rate of falling water level be `- (dh)/(dt)`
Initially at `t =0, h=h`
`t=t,h=0`
Then ` A (- (dh)/(dt)) = pi a ^(2) .v`
`dt = (A)/(pi a ^(2) sqrt (2gh))dh`
`[because` velocity of efflux of liquid `v = sqrt (2gh)]`
Interating both sides
`int _(0) ^(t) dt =- (A)/(sqrt (2) g pi a ^(2)) int _(h) ^(0) h ^(-1//2) dh`
`[t] _(0)^(t) = - (A)/(sqrt(2) g pi a ^(2)) . [ (h ^(1//2) )/( 1//2)] ^(0) implies t = (sqrt2 A )/( pi a ^(2)) sqrt ((h)/(g))`