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

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.

The correct answer is A.