Question

Water is running into an underground right circular conical reservoir, which is $10m$ deep and radius of its base is $5m$. If the rate of change in the volume of water in the reservoir is $\frac{3}{2}\pi m^3/min$, then the rate (in m/min) at which water rises in it, when the water level is $4m$, is:

• A. $\frac{3}{8}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{3}{2}$

Answer 1

Answer: (A)

$\frac{3}{8}$

Depth of cone is $10$ m and its radius is $5$ m.

Rate of change of volume of water is $\frac{3}{2}\pi$.

WE need to find rate at which water risesin it, when water level is $4$ m.

$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi h^3{\tan }^2\theta$

$\frac{du}{dt}=\pi r^2\frac{dh}{dt}$

Now $\frac{r}{4}=\frac{5}{10}\Rightarrow r=2$

$\Rightarrow \frac{3}{2}\pi =x(2{)}^2\frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{3}{8}$