If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is (g = acceleration due to gravity near the poles and R is the radius of earth) (Ignore equatorial bulge)

Options

A. `2 pi sqrt((2R)/(g))`
B. `2pi sqrt((R)/(2g))`
C. `2 pi sqrt((R)/(3g))`
D. `2 pi sqrt((R)/(g))`

Explanation

To solve the problem, we need to determine the time period of rotation of the Earth if the weight of a person at the equator becomes half of their weight at the poles.

### Step-by-Step Solution:

1. **Understanding 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 poles.
- At the equator, due to the Earth's rotation, the effective weight \( W_e \) is given by \( W_e = mg - mR\omega^2 \), where \( R \) is the radius of the Earth and \( \omega \) is the angular velocity of the Earth.

2. **Setting 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 weight, we have:
\[ mg - mR\omega^2 = \frac{1}{2} mg \]

3. **Simplifying the Equation**:
- Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ g - R\omega^2 = \frac{1}{2} g \]
- Rearranging gives:
\[ g - \frac{1}{2}g = R\omega^2 \]
\[ \frac{1}{2}g = R\omega^2 \]

4. **Solving for Angular Velocity**:
- Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{g}{2R} \]
- Taking the square root gives us:
\[ \omega = \sqrt{\frac{g}{2R}} \]

5. **Finding the Time Period**:
- The time period \( T \) of rotation is related to angular velocity \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substituting the expression for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{\frac{g}{2R}}} \]
- This simplifies to:
\[ T = 2\pi \sqrt{\frac{2R}{g}} \]

6. **Final Answer**:
- Thus, the time period of rotation of the Earth, under the given conditions, is:
\[ T = 2\pi \sqrt{\frac{2R}{g}} \]