A cylindrical drum of radius 7 cm and height 2 m is being kept in a vertical position filled with milk. If the milk is leaking at \(14\ \mathrm{cm^3/sec}\) from its lower base, then the rate of decrease in the level of milk is: [Take \(\pi = \displaystyle\frac{22}{7}\)]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{11}\) cm/sec
Let h cm be the level of milk at time t. The radius r = 7 cm stays fixed, so the volume of milk is
\(V = \pi r^2 h = \pi (7)^2 h = 49\pi h\)
Differentiating with respect to t (related rates):
\( \displaystyle \frac{dV}{dt} = 49\pi \frac{dh}{dt} \)
The milk leaks at 14 cm3/sec, so the volume falls at that rate:
\( \displaystyle 14 = 49 \times \frac{22}{7} \times \left(-\frac{dh}{dt}\right) = 154\left(-\frac{dh}{dt}\right) \)
\( \displaystyle -\frac{dh}{dt} = \frac{14}{154} = \frac{1}{11} \)
So the level of milk decreases at \(\frac{1}{11}\) cm/sec. The height of the drum (2 m) is not needed.