A liquid kept in a cylindrical vessel of radius 0.3 m is rotated with a speed 2 r.p.s. The difference in the height of the liquid at the centre of the vessel and at it's sides it

Options

A. 0.01 m
B. 0.02 m
C. 0.04 m
D. 0.8 m

Explanation

To solve the problem of finding the difference in height of the liquid at the center of a cylindrical vessel and at its sides when the vessel is rotated, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Radius of the cylindrical vessel, \( R = 0.3 \, \text{m} \)
- Speed of rotation, \( n = 2 \, \text{r.p.s} \)

2. **Convert Revolutions per Second to Angular Velocity**:
- The angular velocity \( \omega \) in radians per second is given by:
\[ \omega = 2 \pi n = 2 \pi \times 2 = 4 \pi \, \text{rad/s} \]

3. **Calculate the Linear Velocity at the Radius**:
- The linear velocity \( V \) at the edge of the vessel is given by:
\[ V = R \omega = 0.3 \times 4\pi = 1.2\pi \, \text{m/s} \]

4. **Apply Bernoulli's Equation**:
- According to Bernoulli's principle, we can relate the pressure difference between the center and the side of the liquid. The difference in height \( H \) can be calculated using:
\[ H = \frac{V^2}{2g} \]
- Where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

5. **Substitute the Values**:
- First, calculate \( V^2 \):
\[ V^2 = (1.2\pi)^2 = 1.44\pi^2 \, \text{m}^2/\text{s}^2 \]
- Now, substitute into the height equation:
\[ H = \frac{1.44\pi^2}{2 \times 10} = \frac{1.44\pi^2}{20} = 0.072\pi^2 \, \text{m} \]

6. **Calculate the Numerical Value**:
- Using \( \pi \approx 3.14 \):
\[ H \approx 0.072 \times (3.14)^2 \approx 0.072 \times 9.8596 \approx 0.710 \, \text{m} \]

7. **Final Result**:
- The difference in height of the liquid at the center and at the sides of the vessel is approximately:
\[ H \approx 0.71 \, \text{m} \]

### Conclusion:
The difference in the height of the liquid at the center of the vessel and at its sides is approximately **0.71 meters**.