Question

A cylindrical tank of height 0.4 m is open at the top and has a diameter of 0.16 m. It is filled with water upto a height of 0.16 m. Find the time taken to empty the tank through a hole of radius 5 x 10^-3 m in its bottom.

• A. t = 23.5 s
• B. t = 47.3 s
• C. t = 19.02 s
• D. t = 46.26 s

Answer 1

The correct option is D

t = 46.26 s

The velocity of efflux through a small hole at a depth h is given by
$v=\sqrt{2gh}.$
According to the principle of continuity,
We get,$-\pi R^2\frac{dh}{dt}=\pi r^{2}v.$ Since the rate of fall of water level in the tank is - $\frac{dh}{dt}$
$\Rightarrow -R^2\frac{dh}{dt}=r^2\sqrt{2gh}$
$\Rightarrow -\frac{dh}{\sqrt{h}}=\frac{r^2}{R^2}\sqrt{2g.}dt.$
Integrating we get,$-{\int }_{0.16}^0\frac{dh}{\sqrt{h}}={\left(\frac{0.005}{0.08}\right)}^2\sqrt{2\times 9.8}{\int }_0^{t}dt$
Therefore, t = 46.26 sec.