A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is `r` and the speed of revolution is `n` rotations/second, find the difference in height of the liquid at the centre of vessel and its sides.

Options

A. `h=(2pi^(2)n^(2)r^(2))/g`
B. `h=(4pi^(2)n^(2)r^(2))/g`
C. `h=(pi^(2)r^(2)n^(2))/g`
D. `h=(pi^(2)r^(2)n^(2))/g`

Explanation

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

### Step-by-Step Solution:

1. **Understand the System**:
- We have a cylindrical vessel of radius \( r \) that is rotating about its axis at a speed of \( n \) rotations per second.
- The liquid inside the vessel will experience a centrifugal force due to the rotation, causing it to rise at the sides of the vessel.

2. **Convert Rotational Speed to Angular Speed**:
- The angular speed \( \omega \) in radians per second can be calculated from the rotational speed \( n \):
\[ \omega = 2 \pi n \quad \text{(since one rotation is } 2\pi \text{ radians)} \]

3. **Consider the Pressure Variation**:
- The pressure at a point in the liquid at a distance \( x \) from the axis of rotation can be expressed considering the centrifugal force. The pressure at the center (point O) is atmospheric pressure \( P_{\text{atm}} \).
- As we move outward to a point at distance \( x \), the pressure increases due to the centrifugal effect.

4. **Set Up the Pressure Equation**:
- For a small height \( dy \) at distance \( x \), the pressure at that point can be expressed as:
\[ P_B = P_{\text{atm}} + \rho \omega^2 x dx - \rho g dy \]
- Here, \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( dy \) is the small height change.

5. **Apply Hydrostatic Equilibrium**:
- In hydrostatic equilibrium, the change in pressure must equal zero:
\[ \rho \omega^2 x dx - \rho g dy = 0 \]
- This leads to:
\[ \omega^2 x dx = g dy \]

6. **Integrate to Find Height Change**:
- Rearranging gives:
\[ dy = \frac{\omega^2 x}{g} dx \]
- Integrate both sides from the center (0) to the radius \( r \):
\[ \int_0^h dy = \int_0^r \frac{\omega^2 x}{g} dx \]
- The left side integrates to \( h \), and the right side can be calculated:
\[ h = \frac{\omega^2}{g} \cdot \left[ \frac{x^2}{2} \right]_0^r = \frac{\omega^2 r^2}{2g} \]

7. **Substitute for Angular Speed**:
- Substitute \( \omega = 2 \pi n \):
\[ h = \frac{(2 \pi n)^2 r^2}{2g} = \frac{4 \pi^2 n^2 r^2}{2g} = \frac{2 \pi^2 n^2 r^2}{g} \]

8. **Determine Height Difference**:
- The height at the center of the vessel is \( 0 \), and at the sides, it is \( h \). Therefore, the difference in height \( \Delta h \) is:
\[ \Delta h = h - 0 = \frac{2 \pi^2 n^2 r^2}{g} \]

### Final Answer:
The difference in height of the liquid at the center of the vessel and its sides is:
\[ \Delta h = \frac{2 \pi^2 n^2 r^2}{g} \]