A cylinderical tank of height 0.4 m is open at the top and has a diameter 0.16m. Water is filled in it up to height of 0.16 m. Find the time taken to empty the tank through a hole of radius `5 xx 10^(-3)m` in its bottom.

Explanation

The velocity of efflux through the hole at a depth h from the free surface of water in tank is : `upsilon=sqrt(2gh)`. If R, r are the radius of cylindrical tank and hole respectively and dh is the decresse in height of water in time dt due to flow of water through hole, then, using the principle of continuity at the top and hole, we have
`-pi R^(2)(dh)/(dt) = pi r^(2)upsilon = pi r^(2)sqrt(2gh)`
or `(-dh)/(dt) = (r^2)/(R^2) sqrt(2g)dt`
Intergrating it within the conditions of given problem, we have
`-int_(h)^(0) = ((5 xx 10^(-3))/(0.08))^(2) sqrt(2 xx 9.8) int_(0)^(t) dt`
`2sqrt(h) = (0.005/0.08)^(2) sqrt(2 xx 9.8) xxt`
or, `t = (2sqrt(0.16))/ (((0.005)/(0.08))^(2)) =sqrt(2 xx 9.8) = 46.26 s`.