Question

A hemispherical tank of radius $R$ is completely filled with water. Now an orifice of small area '$a$' is made at the bottom of tank. The time required to empty the tank is

• A. $\frac{14\pi R^{\frac{5}{2}}}{15\sqrt{2}a\sqrt{g}}$
• B. $\frac{14\pi R^{\frac{5}{2}}}{\sqrt{2}a\sqrt{g}}$
• C. $\frac{14\pi R^{\frac{5}{2}}}{5\sqrt{2}a\sqrt{g}}$
• D. $\frac{4\pi R^{\frac{5}{2}}}{15\sqrt{2}a\sqrt{g}}$

Answer 1

The correct option is A $\frac{14\pi R^{\frac{5}{2}}}{15\sqrt{2}a\sqrt{g}}$

Differenciating volume of tank w.r.t height of tank

$dV=\pi x^{2}dh$

Since $x^2=R^2-(R-h{)}^2$

$=\pi (2Rh-h^2)dh$

Volume water coming out of orifince w.r.t time

$-\frac{dV}{dt}=a\sqrt{2gh}$

Substituting values

$ıdt=l\frac{-\pi }{a\sqrt{2g}}(2R{h}^{\frac{1}{2}}-h^{\frac{3}{2}})dh$

$t=\frac{14\pi R^{\frac{5}{2}}}{15\sqrt{2}a\sqrt{9}}$