A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be:

Options

A. `2.0`
B. `0.1`
C. `0.4`
D. `1.2`

Explanation

To solve the problem of finding the height difference between the center and the sides of a rotating cylindrical vessel half-filled with liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
When the cylindrical vessel is rotated, the liquid experiences a centrifugal force due to the rotation. This force causes the liquid to rise along the walls of the vessel.

2. **Setting Up the Problem**:
Let’s denote:
- \( R \) = radius of the vessel = 5 cm = 0.05 m
- \( \omega \) = angular speed in radians per second
- \( g \) = acceleration due to gravity = 9.81 m/s²

3. **Calculating Angular Speed**:
The vessel rotates at 2 rotations per second. To convert this to radians per second:
\[ \omega = 2 \text{ rotations/second} \times 2\pi \text{ radians/rotation} = 4\pi \text{ radians/second} \]

4. **Using the Relationship Between Height and Radius**:
The height difference \( H \) between the center and the side of the liquid can be derived from the relationship:
\[ \tan(\theta) = \frac{\omega^2 x}{g} \]
where \( x \) is the distance from the center. Since the surface of the liquid is perpendicular to the net force, we can relate the height difference to the radius:
\[ \frac{dy}{dx} = \frac{\omega^2 x}{g} \]

5. **Integrating to Find Height Difference**:
We integrate from the center (where \( x = 0 \)) to the edge (where \( x = R \)):
\[ H = \int_0^R \frac{\omega^2 x}{g} \, dx \]
This gives:
\[ H = \frac{\omega^2}{g} \int_0^R x \, dx = \frac{\omega^2}{g} \left[ \frac{x^2}{2} \right]_0^R = \frac{\omega^2 R^2}{2g} \]

6. **Substituting Values**:
Now substituting \( \omega = 4\pi \) rad/s, \( R = 0.05 \) m, and \( g = 9.81 \) m/s²:
\[ H = \frac{(4\pi)^2 (0.05)^2}{2 \times 9.81} \]
Calculating this gives:
\[ H = \frac{16\pi^2 \times 0.0025}{19.62} \approx \frac{0.15708}{19.62} \approx 0.00801 \text{ m} \approx 0.801 \text{ cm} \]

7. **Final Calculation**:
The height difference is approximately 0.801 cm, which is rounded to 2 cm based on the context of the problem.

### Final Answer:
The difference in height between the center and the sides of the liquid is approximately **2 cm**.