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 $14\, cm^3/sec$ from its lower base, then the rate of decrease in the level of milk is: [Take $π =\frac{22}{7}$]

$\frac{1}{11}$ cm/sec

The correct answer is Option (2) → $\frac{1}{11}$ cm/sec

Radius of cylinder: $r = 7 \text{ cm}$

Height: $h = 2 \text{ m} = 200 \text{ cm}$

Volume of cylinder: $V = \pi r^2 h$

Given: Rate of leakage $= \frac{dV}{dt} = 14 \text{ cm}^3/\text{sec}$

Let $h$ be the height of milk at time $t$. Then $V = \pi r^2 h$

Differentiating: $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$

$14 = \frac{22}{7} \cdot 7^2 \cdot \frac{dh}{dt}$

$14 = 154 \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{14}{154} = \frac{1}{11} \text{ cm/sec}$

Rate of decrease in milk level = $\frac{1}{11} \text{ cm/sec}$