An inverted conical flask is being filled with water at the rate of...

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

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

$⇒ \dfrac{h}{10} = \dfrac{r}{5}$

$⇒ r = \dfrac{5h}{10}$

$⇒ r = \dfrac{h}{2}$

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

$⇒ V = \dfrac{\pih^{3}}{12}$

On differentiating both sides, we get

$\dfrac{dV}{dt} = \dfrac{\pi}{12} \cdot 3h^{2}\dfrac{dh}{dt} = \dfrac{\pih^{2}}{4}\dfrac{dh}{dt}$

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

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

$⇒ \dfrac{dh}{dt} = \dfrac{3}{4\pi}cm\sec^{-1}$

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

$∴$ Option (b) is correct.