If a man at the equator would weigh `(3//5)`th of his weight, the angular speed of the earth is

Options

A. `sqrt(2/5 g/R)`
B. `sqrt(g/R)`
C. `sqrt(R/g)`
D. `sqrt(2/5 R/g)`

Explanation

To solve the problem, we need to find the angular speed of the Earth given that a man at the equator weighs \( \frac{3}{5} \) of his actual weight due to the centrifugal force acting on him due to the Earth's rotation.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- At the equator, a man experiences two forces:
- The gravitational force acting downward, which is \( mg \) (where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity).
- The centrifugal force acting outward due to the Earth's rotation, which is given by \( m \omega^2 r \) (where \( \omega \) is the angular speed of the Earth and \( r \) is the radius of the Earth).

2. **Setting Up the Equation**:
- According to the problem, the effective weight of the man at the equator is \( \frac{3}{5} \) of his actual weight. Therefore, we can write the equation:
\[ \frac{3}{5} mg = mg - m \omega^2 r \]
- Here, \( \frac{3}{5} mg \) is the apparent weight, and \( mg - m \omega^2 r \) is the net force acting on the man.

3. **Simplifying the Equation**:
- We can cancel \( m \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ \frac{3}{5} g = g - \omega^2 r \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ \omega^2 r = g - \frac{3}{5} g \]
- This simplifies to:
\[ \omega^2 r = \frac{2}{5} g \]

5. **Solving for Angular Speed**:
- Now, we can solve for \( \omega^2 \):
\[ \omega^2 = \frac{2g}{5r} \]
- Taking the square root gives:
\[ \omega = \sqrt{\frac{2g}{5r}} \]

6. **Final Expression**:
- Thus, the angular speed of the Earth is:
\[ \omega = \sqrt{\frac{2g}{5r}} \]

### Answer:
The angular speed of the Earth is \( \sqrt{\frac{2g}{5r}} \).