A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lower most. Its semi vertical angle is \( \tan^{-1}\left(\displaystyle\frac{2}{3}\right) \). Water is poured into it at a constant rate 6 cubic metre per hour. The rate at which the level of the water is rising at the instant when the depth of water in the tank is 5 meter is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{27}{50\pi} \) meter/hour
Let r be the radius of the water surface and h the depth. The semi vertical angle \( \alpha \) gives
\( \displaystyle \tan\alpha=\frac{r}{h}=\frac{2}{3} \)
\( \displaystyle r=\frac{2h}{3} \)
So the volume can be written in h alone:
\( \displaystyle V=\frac{1}{3}\pi r^2h \)
\( \displaystyle =\frac{1}{3}\pi\cdot\frac{4h^2}{9}\cdot h \)
\( \displaystyle =\frac{4\pi h^3}{27} \)
Differentiating with respect to t:
\( \displaystyle \frac{dV}{dt}=\frac{4\pi h^2}{9}\cdot\frac{dh}{dt} \)
With \( \frac{dV}{dt}=6 \) and h = 5:
\( \displaystyle 6=\frac{4\pi(25)}{9}\cdot\frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt}=\frac{54}{100\pi}=\frac{27}{50\pi} \)
The level rises at \( \frac{27}{50\pi} \) meter/hour.
NCERT (2026-27 edition):
Rate of Change of Quantities
