A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25 π m³/minute. The rate at which the surface of the oil is rising, is

1 m/minute

Let r be the radius, h be the height and V be the volume of the oil at time t. Then,

$V=\pi r^2 h$

$\Rightarrow V=\frac{\pi}{4} h$                [∵ r = 0.5 m = $\frac{1}{2}$ (given)]

$\Rightarrow \frac{d V}{d t}=\frac{\pi}{4} \frac{d h}{d t}$

$\Rightarrow 0.25 \pi=\frac{\pi}{4} \times \frac{d h}{d t}$         [∵ $\frac{d V}{d t}=0.25 \pi$ m³ (given)]  

$\Rightarrow \frac{d h}{d t}$ = 1 m/minute