There is a hole of area `(1)/(25)cm^(2)` in the bottom of a cylindrical vessel containing fluid up to height `h`. The liquid flows out in time `t`. If the liquid were filled in the vessel up to height `4h`, then it would flow out in time

Options

A. t
B. 2t
C. 4t
D. `(t)/(2)`

Answer

B

Explanation

Let `A` and `a` be the corss-sectional areas of the vesser and hole respectively. Let `h` be the height of water in the vessel at time. Let `(-(dh)/(dt))` represent the rate of fall of level.
then `A(-(dh)/(dt)) = alpha v = asqrt(2gh)`
or `-(dh)/(dt) = (alphasqrt(2g))/(A) dt`
`-int_(A)^(0) (1)/(sqrt(h))dh = (asqrt(2g))/(A) int_(0)^(g) dt`
`-(-2sqrt(h)) = (alphasqrt(2g))/(A)t`
or `t=(A)/(alpha) (1)/(sqrt(2g)) xx 2sqrt(h) or t = A/(alpha) sqrt((2h)/(g))`
Now `t prop sqrt(h)`
where `h` is quarupled, `t` is doubled.