A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the side. If the radius of the vessel is `5 cm` and the speed of rotation is `4 rev//s`, then the difference in the height of the liquid at the centre of the vessel and its sides is

Options

A. (a)8 cm
B. (b)2 cm
C. (c)40 cm
D. (d)4 cm

Explanation

To solve the problem, we need to determine the difference in height of the liquid at the center of the cylindrical vessel and at its sides when the vessel is rotated. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Radius of the vessel, \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Speed of rotation, \( f = 4 \, \text{rev/s} \)

### Step 2: Convert the speed of rotation to angular velocity
Angular velocity \( \omega \) in radians per second can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the given value:
\[ \omega = 2\pi \times 4 = 8\pi \, \text{rad/s} \]

### Step 3: Apply Bernoulli's equation
We apply Bernoulli's equation between the center of the vessel and the side of the vessel. The pressure difference can be expressed as:
\[ P_1 - P_2 = \frac{1}{2} \rho v^2 \]
Where \( P_1 \) is the pressure at the center, \( P_2 \) is the pressure at the side, \( \rho \) is the density of the liquid, and \( v \) is the linear velocity at the side of the vessel.

### Step 4: Calculate the linear velocity at the side
The linear velocity \( v \) at the side of the vessel is given by:
\[ v = r \omega \]
Substituting the values:
\[ v = 0.05 \times 8\pi = 0.4\pi \, \text{m/s} \]

### Step 5: Substitute into Bernoulli's equation
The pressure difference can now be expressed as:
\[ P_1 - P_2 = \frac{1}{2} \rho (0.4\pi)^2 \]

### Step 6: Relate pressure difference to height difference
The pressure difference can also be expressed in terms of height difference \( h \):
\[ P_1 - P_2 = \rho g h \]
Equating the two expressions for pressure difference:
\[ \rho g h = \frac{1}{2} \rho (0.4\pi)^2 \]
The density \( \rho \) cancels out:
\[ g h = \frac{1}{2} (0.4\pi)^2 \]

### Step 7: Solve for height difference \( h \)
Rearranging gives:
\[ h = \frac{(0.4\pi)^2}{2g} \]
Substituting \( g \approx 10 \, \text{m/s}^2 \):
\[ h = \frac{(0.4\pi)^2}{20} \]

### Step 8: Calculate the value of \( h \)
Calculating \( (0.4\pi)^2 \):
\[ (0.4\pi)^2 = 0.16\pi^2 \]
Thus,
\[ h = \frac{0.16\pi^2}{20} = 0.008\pi^2 \, \text{m} \]
Calculating \( \pi^2 \approx 9.87 \):
\[ h \approx 0.008 \times 9.87 \approx 0.079 \, \text{m} \approx 8 \, \text{cm} \]

### Final Answer
The difference in height of the liquid at the center of the vessel and its sides is approximately **8 cm**.

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