Find the value of angular velocity of axial rotation of the earth, such that weight of a person at equator becomes 3/4 of its weight at pole, Radius of the earth at equator is 6400 km.

Options

A. `62xx10^(-4)"rads"^(-1)`
B. `52xx10^(-4)"rads"^(-1)`.
C. `62xx10^(-2)"rads"^(-1)`.
D. NONE OF THE ABOVE

Explanation

To find the angular velocity of the Earth's axial rotation such that the weight of a person at the equator becomes 3/4 of their weight at the pole, 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 \( W_{\text{pole}} = mg \), where \( g \) is the acceleration due to gravity at the poles.
- The weight of a person at the equator is given by \( W_{\text{equator}} = m(g - R\omega^2) \), where \( R \) is the radius of the Earth and \( \omega \) is the angular velocity.

2. **Set Up the Equation**:
- According to the problem, the weight at the equator is \( \frac{3}{4} \) of the weight at the poles:
\[ W_{\text{equator}} = \frac{3}{4} W_{\text{pole}} \]
- Substituting the expressions for weight, we have:
\[ m(g - R\omega^2) = \frac{3}{4} mg \]

3. **Cancel the Mass**:
- Since mass \( m \) is common on both sides, we can cancel it out:
\[ g - R\omega^2 = \frac{3}{4} g \]

4. **Rearrange the Equation**:
- Rearranging gives:
\[ g - \frac{3}{4} g = R\omega^2 \]
- Simplifying this leads to:
\[ \frac{1}{4} g = R\omega^2 \]

5. **Solve for Angular Velocity \( \omega \)**:
- Rearranging for \( \omega \):
\[ \omega^2 = \frac{g}{4R} \]
\[ \omega = \sqrt{\frac{g}{4R}} \]

6. **Substitute Values**:
- Given \( g = 9.8 \, \text{m/s}^2 \) and \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \):
\[ \omega = \sqrt{\frac{9.8}{4 \times 6400 \times 10^3}} \]

7. **Calculate \( \omega \)**:
- First calculate \( 4R \):
\[ 4R = 4 \times 6400 \times 10^3 = 25600 \times 10^3 = 2.56 \times 10^7 \, \text{m} \]
- Now substitute this back into the equation for \( \omega \):
\[ \omega = \sqrt{\frac{9.8}{2.56 \times 10^7}} \approx \sqrt{3.84375 \times 10^{-7}} \approx 0.000620 \, \text{radians/s} \]

8. **Final Result**:
- Thus, the angular velocity \( \omega \) is approximately:
\[ \omega \approx 6.2 \times 10^{-4} \, \text{radians/s} \]