The angular speed of rotation of earth about its axis at which the weight of man standing on the equator becomes half of his weight at the poles is given by

Options

A. `0.034 rad s^(-1)`
B. `8.75 xx 10^(-4) rad s^(-1)`
C. `1.23 xx 10^(-2) rad s^(-1)`
D. `7.65 xx 10^(-7) rad s^(-1)`

Explanation

To solve the problem of finding the angular speed of rotation of the Earth about its axis at which the weight of a man standing on the equator becomes half of his weight at the poles, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Weight at the Poles and Equator**:
- The weight of a person at the poles is given by \( W_p = mg \), where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity.
- At the equator, the effective weight \( W_e \) is reduced due to the centrifugal force caused by the Earth's rotation. This can be expressed as:
\[ W_e = mg - m \omega^2 r \]
- Here, \( \omega \) is the angular speed of the Earth, and \( r \) is the radius of the Earth.

2. **Set Up the Equation**:
- According to the problem, the weight at the equator is half of the weight at the poles:
\[ W_e = \frac{1}{2} W_p \]
- Substituting the expressions for \( W_e \) and \( W_p \):
\[ mg - m \omega^2 r = \frac{1}{2} mg \]

3. **Simplify the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g - \omega^2 r = \frac{1}{2} g \]
- Rearranging gives:
\[ g - \frac{1}{2} g = \omega^2 r \]
- Thus:
\[ \frac{1}{2} g = \omega^2 r \]

4. **Solve for Angular Speed \( \omega \)**:
- Rearranging the equation gives:
\[ \omega^2 = \frac{g}{2r} \]
- Taking the square root:
\[ \omega = \sqrt{\frac{g}{2r}} \]

5. **Substitute Known Values**:
- Using \( g \approx 9.8 \, \text{m/s}^2 \) and \( r \approx 6400 \times 10^3 \, \text{m} \):
\[ \omega = \sqrt{\frac{9.8}{2 \times 6400 \times 10^3}} \]

6. **Calculate \( \omega \)**:
- First calculate \( 2r \):
\[ 2r = 2 \times 6400 \times 10^3 = 12800 \times 10^3 \]
- Now substitute into the equation:
\[ \omega = \sqrt{\frac{9.8}{12800 \times 10^3}} \approx \sqrt{7.65625 \times 10^{-6}} \approx 0.00277 \, \text{rad/s} \]

7. **Final Result**:
- Therefore, the angular speed at which the weight of a man standing on the equator becomes half of his weight at the poles is approximately:
\[ \omega \approx 2.77 \times 10^{-3} \, \text{rad/s} \]