From a cylindrical drum containing petrol and kept vertical, the petrol is leaking at the rate of $10\, cm^3/sec$. If the radius of the drum is 25 cm and height 1 metre, find the rate at which the level of the petrol is changing when the petrol level is 80 cm.

$\frac{2}{125\pi} \, \text{cm/sec}$

The correct answer is Option (1) → $\frac{2}{125\pi} \, \text{cm/sec}$

When the petrol from the cylindrical drum leaks, then only the height of the petrol will change while its radius remains constant at 25 cm.

Let h be the height of the petrol in the drum at any time t and V be its volume, then

$V = π × (25)^2 × h = 625 πh$, diff. w.r.t. t, we get

$\frac{dV}{dt}= 625 π.\frac{dh}{dt}$

Since the petrol is leaking at the rate of $10\, cm^3/sec, \frac{dV}{dt}= -10\, cm^3/sec$

($\frac{dV}{dt}$ is -ve, because V is decreasing)

$⇒-10=625π\frac{dh}{dt}⇒\frac{dh}{dt}=-\frac{2}{125\pi}$, which is constant.

Hence, the level of the petrol is decreasing at the rate of $\frac{2}{125\pi} \, \text{cm/sec}$ when the petrol level is 80 cm.