A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of(A) 1 `m//h` (B) 0.1 `m^3//h` (C) 1.1 `m^3//h` (D) 0.5 `m^3//h`

Options

A. 1 m/h
B. `0.1 m//h`
C. ` 1.1 m//h`
D. `0.5 m//h`

Explanation

To solve the problem, we need to find the rate at which the depth of wheat in a cylindrical tank is increasing when the tank is being filled at a certain volume rate.

### Step-by-Step Solution:

1. **Identify the given values**:
- Radius of the cylindrical tank, \( r = 10 \) m
- Rate at which the tank is being filled, \( \frac{dV}{dt} = 314 \) m³/h

2. **Write the formula for the volume of a cylinder**:
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( h \) is the height (or depth) of the wheat.

3. **Substitute the radius into the volume formula**:
Since the radius \( r = 10 \) m, we can substitute this into the volume formula:
\[ V = \pi (10)^2 h = 100\pi h \]

4. **Differentiate the volume with respect to time \( t \)**:
We differentiate both sides of the volume equation with respect to \( t \):
\[ \frac{dV}{dt} = 100\pi \frac{dh}{dt} \]

5. **Substitute the known rate of volume change**:
We know that \( \frac{dV}{dt} = 314 \) m³/h. Therefore, we can substitute this value into the differentiated equation:
\[ 314 = 100\pi \frac{dh}{dt} \]

6. **Solve for \( \frac{dh}{dt} \)**:
Rearranging the equation to solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{314}{100\pi} \]

7. **Calculate \( \frac{dh}{dt} \)**:
Using \( \pi \approx 3.14 \):
\[ \frac{dh}{dt} = \frac{314}{100 \times 3.14} \approx \frac{314}{314} = 1 \text{ m/h} \]

### Conclusion:
The depth of the wheat is increasing at the rate of **1 m/h**. Therefore, the correct option is (A) 1 m/h.