Find the time required for a cylindrical tank of radius `r` and height `H` to empty through a round hole of area `a` at the bottom. The flow through the hole is according to the law `v(t)=ksqrt(2gh(t))` , where `v(t)` and `h(t)` , are respectively, the velocity of flow through the hole and the height of the water level above the hole at time `t ,` and `g` is the acceleration due to gravity.

Explanation

Let at time t the depth of water is h and radius of water surface is r. If in time dt the decrease of water level is dh, then
`-pir^(2)dh=aksqrt(2gh)dt`
or `(-pir^(2))/(aksqrt(2h)sqrt(h))=dt`
or `(-pir^(2))/(aksqrt(2g))=(dh)/sqrt(h)=dt`
Now when t=0, h=H and when t=t, h=0, then
`-(pir^(2))/(aksqrt(2g))int_(H)^(0)(dh)/sqrt(h) = int_(0)^(t)dt`
or `(-pir^(2))/(aksqrt(2g)){2sqrt(h)}_(H)^(0)=t`
or `t=(pir^(2)2sqrtH)/(aksqrt(2g)) = (pir^(2))/(ak) sqrt((2H)/(g))`