A cylindrical tank of height 0.4m is open at the top and has a diameter 0.16m. Water is filled in it uo to height of 0.16m. Find the time taken to empty the tank through a hole of radius `5xx10^(-3)m` in its bottom.

Explanation

The velocity of efflux through the hole `V=sqrt2gh`
Let R, r nr the radius of cylindrical tank and hole and .dh. is the geight of water in the .dt.sec use principle of continuity at the top and hole .
then `-piR^(2)(dh)/(dt)=pir^(2)v = pir^(2) sqrt2gh`
or `-(dh)/(sqrth)=(r^(2))/(R^(2))sqrt2g dt`
integrating it wuthin the condition of problem,
`-int_(h)^(0)(dh)/(sqrth)=((5xx10^(-3))/(0.08))^(2) sqrt2xx9.8 int_(0)^(t)dt`
`2sqrth=((0.005)/(0.008))^(2)sqrt2xx9.8xxt`
(or) `t=(2xxsqrt0.16)/(((0.005)/(0.08))^(2) xx sqrt2 xx 9.8) =46.26s`