Water is leaking from bottom of a conical funnel at the rate of \( 0.15\pi\ cm^3/s \) . If the radius of the base of the funnel is 10 cm and its height is 20 cm then the rate at which the water level is dropping when it is 5 cm from the top is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{375}\ cm/s \)
Let the water be h cm deep with surface radius r. By similar triangles, \( \frac{r}{h}=\frac{10}{20} \), so \( r=\frac{h}{2} \):
\( \displaystyle V=\frac{1}{3}\pi\left(\frac{h}{2}\right)^2h \)
\( \displaystyle =\frac{\pi h^3}{12} \)
Differentiating with respect to t:
\( \displaystyle \frac{dV}{dt}=\frac{\pi h^2}{4}\,\frac{dh}{dt} \)
5 cm from the top means h = 20 − 5 = 15, and the volume falls at \( 0.15\pi \) cm3/s:
\( \displaystyle 0.15\pi=\frac{225\pi}{4}\,\frac{dh}{dt} \)
\( \displaystyle \frac{dh}{dt}=\frac{0.6}{225} \)
\( \displaystyle =\frac{1}{375}\ \text{cm/s} \)
NCERT (2026-27 edition):
Application of Derivatives
