A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour.
Question:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour. Then the depth of the wheat is increasing at the rate of

(A) $1 \mathrm{~m} / \mathrm{h}$

(B) $0.1 \mathrm{~m} / \mathrm{h}$

(C) $1.1 \mathrm{~m} / \mathrm{h}$

(D) $0.5 \mathrm{~m} / \mathrm{h}$

Solution:

Let r be the radius of the cylinder.

Then, volume (V) of the cylinder is given by,

\begin{aligned} V &=\pi(\text { radius })^{2} \times \text { height } \\ &=\pi(10)^{2} h \quad(\text { radius }=10 \mathrm{~m}) \\ &=100 \pi h \end{aligned}

Differentiating with respect to time t, we have:

$\frac{d V}{d t}=100 \pi \frac{d h}{d t}$

The tank is being filled with wheat at the rate of 314 cubic metres per hour.

$\therefore \frac{d V}{d t}=314 \mathrm{~m}^{3} / \mathrm{h}$

Thus, we have:

$314=100 \pi \frac{d h}{d t}$

$\Rightarrow \frac{d h}{d t}=\frac{314}{100(3.14)}=\frac{314}{314}=1$

Hence, the depth of wheat is increasing at the rate of 1 m/h.