An inverted conical flask is being filled with water at the rate of $3 {\mathrm{cm}}^3 {\sec }^{-1}$. The height of the flask is $10 \mathrm{cm}$ and the radius of the base is $5 \mathrm{cm}$. How fast is the water level rising when the level is $4 \mathrm{cm}?$

Let depth of water at time $t$ be $h$ and the radius of the base of water level be $r$.

From similar $∆OMP$ and $∆ONQ$, we have

$\frac{OM}{ON}=\frac{PM}{QN}$

$\Rightarrow \frac{h}{10}=\frac{r}{5}$

$\Rightarrow r=\frac{5h}{10}$

$\Rightarrow r=\frac{h}{2}$

We have, $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h$

$\Rightarrow V=\frac{\pi h^3}{12}$

On differentiating both sides, we get

$\frac{dV}{dt}=\frac{\pi }{12}·3h^2\frac{dh}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$

Given, $\frac{dV}{dt}=3 {\mathrm{cm}}^3 {\sec }^{-1}$ when $h=4 \mathrm{cm},$ so we get

$3=\frac{\pi \times 4^2}{4}\frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{3}{4\pi } \mathrm{cm} {\sec }^{-1}$

Hence, the water level is rising at $\frac{3}{4\pi } \mathrm{cm} {\sec }^{-1}$

$\therefore$ Option (b) is correct.