Water is leaking from a conical funnel at the rate of 5 cm^3/sec. If…

The correct answer is Option (2) → $\frac{4}{45\pi}\text{cm/sec}$

At any time t, water forms a cone. Let r cm be the radius of this cone and h cm be its height, then

volume of water in the funnel = $V =\frac{1}{3}πr^2h$.

Since Δs DVB and CVA are similar,

$\frac{DV}{CV}=\frac{DB}{CA}⇒\frac{h}{20}=\frac{r}{10}⇒r=\frac{h}{2}$

∴ Volume of water at any time = V

$=\frac{1}{3}π.(\frac{h}{2})^2.h=\frac{π}{12}h^3$

Diff. w.r.t. t, we get $\frac{dV}{dt}=\frac{\pi}{12}.3h^2\frac{dh}{dt}=\frac{\pi}{4}h^2\frac{dh}{dt}$

Since water is leaking from the conical funnel at the rate of $5\, cm^3/sec,$

$\frac{dV}{dt}=-5\,cm^3/sec$  ($\frac{dV}{dt}$ is -ve, because V is decreasing)

$⇒-5=\frac{\pi}{4}h^2\frac{dh}{dt}⇒\frac{dh}{dt}=-\frac{20}{πh^2}$

When the water level is 5 cm from top, $h = 20-5 = 15$ cm, then

$\frac{dh}{dt}=-\frac{20}{π×(15)^2}=-\frac{4}{45\pi}\text{cm/sec}$

Hence, the rate at which the water level is dropping =$\frac{4}{45\pi}\text{cm/sec}$