Question

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

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

Answer 1

The correct option is D $\frac{3}{8}$

Let $h$ be the height of water and $r$ be the radius of right circular conical reservoir at time $t.$


$△OAB$ is similar to $△OCD$
So, $\frac{AB}{OB}=\frac{CD}{OD}$
$\Rightarrow \frac{5}{10}=\frac{r}{h}$
$\Rightarrow r=\frac{h}{2}$
Volume of water in right circular conical reservoir, $V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow V=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h$
$\Rightarrow V=\frac{\pi }{12}{h}^3$
Differentiating w.r.t. $t$, we get
$\frac{dV}{dt}=\frac{\pi }{4}\times h^2\times \frac{dh}{dt}$
Given, $\frac{dV}{dt}=\frac{3\pi }{2}$ and $h=4$
$\Rightarrow \frac{3\pi }{2}=\frac{\pi }{4}\times 4^2\times \frac{dh}{dt}$
$\therefore \frac{dh}{dt}=\frac{3}{8}\text{m}/\text{min}$