Water is being poured into a conical tank of height $12 m$ and radius $4 m$ . The water level rises at a constant rate of $0.5 m/min$ . Find the rate of increase of the volume of water when the water level is $6 m$ .

hard 4 marks Class 12 · Mathematics · Application of Derivatives

Answer

2 π m^{3} / min

The correct answer is $2 π m^{3} / min$ — Thus the volume increases at $2 π$ cubic metres per minute when the water level is $6 m$ .

Step-by-step solution

The radius $r$ of the water surface is related to the height $h$ by similar triangles: $r / h = 4/12 = 1/3$ , so $r = h /3$ .
Volume of water in a cone: $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π! (\frac{h}{3})^{2} h = \frac{π}{27} h^{3}$ .
Differentiate implicitly with respect to time: $\frac{d V}{d t} = \frac{π}{9} h^{2} \frac{d h}{d t}$ .
Substitute $h = 6 m$ and $\frac{d h}{d t} = 0.5 m/min$ : $\frac{d V}{d t} = \frac{π}{9} \cdot 36 \cdot 0.5 = \frac{π}{9} \cdot 18 = 2 π m^{3} / min$ .
Thus the volume increases at $2 π$ cubic metres per minute when the water level is $6 m$ .

Common mistake: Forgetting to express the radius in terms of the height

Concept tested: Related rates for a conical tank

Published 2 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked