Question

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) = k $\sqrt{2gh(t)}$ 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 –

• A. $\frac{\pi r^{2}}{ak}\sqrt{\left( \frac{H}{g} \right)}$
• B. $\frac{\pi r^{2}}{ak}\sqrt{\left( \frac{2H}{g} \right)}$
• C. $\frac{\pi r^{3}}{ak}\sqrt{\left( \frac{2H}{g} \right)}$
• D. None

Answer 1

Answer: (B)

$\frac{\pi r^{2}}{ak}\sqrt{\left( \frac{2H}{g} \right)}$

Answer 2

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

– pr² dh = ak $\sqrt{(2gh)}$dt

Ž $\frac{–\pi r^{2}}{ak\sqrt{2g}\sqrt{h}}$dh = dt

Ž – $\frac{–\pi r^{2}}{ak\sqrt{2g}}\frac{dh}{\sqrt{h}}$ = dt

Now when t = 0, h = H

and when t = t, h = 0

then $\int_{H}^{0}\frac{dh}{\sqrt{h}}$=$\int_{0}^{t}{dt}$

Ž $\left\{ 2\sqrt{h} \right\}_{H}^{0}$ = t

Ž = $\frac{\pi r^{2}.2\sqrt{H}}{ak\sqrt{2g}}$ = $\frac{\pi r^{2}}{ak}$ $\sqrt{\left( \frac{2H}{g} \right)}$