Water is running into a conical vessel, 18cm deep and 6cm in radius, at the rate of 0.1 cm3/sec. If the water is 4cm deep, then the water level rising at the rate
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{9}{160\pi} \) cm/sec
Let the water have depth h and surface radius r. By similar triangles, \( \frac{r}{h}=\frac{6}{18} \), so \( r=\frac{h}{3} \).
\( \displaystyle V=\frac{1}{3}\pi r^2h \)
\( \displaystyle =\frac{1}{3}\pi\cdot\frac{h^2}{9}\cdot h=\frac{\pi h^3}{27} \)
\( \displaystyle \frac{dV}{dt}=\frac{\pi h^2}{9}\cdot\frac{dh}{dt} \)
Putting \( \frac{dV}{dt}=0.1 \) and h = 4:
\( \displaystyle 0.1=\frac{16\pi}{9}\cdot\frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt}=\frac{0.9}{16\pi}=\frac{9}{160\pi}\ \text{cm/sec} \)
NCERT (2026-27 edition):
Rate of Change of Quantities
