The speed of the earth's rotation about it's axis is `omega`. Its angular speed is increased to make the effective acceleration due to gravity equal to zero at the equator. Then what will be the final `omega`.

Options

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

Explanation

To solve the problem, we need to find the final angular speed \( \omega_1 \) of the Earth such that the effective acceleration due to gravity at the equator becomes zero.

### Step-by-step Solution:

1. **Understanding Effective Gravity**:
The effective acceleration due to gravity \( g_{\text{effective}} \) at the equator can be expressed as:
\[ g_{\text{effective}} = g - r \omega^2 \]
where \( g \) is the standard acceleration due to gravity, \( r \) is the radius of the Earth, and \( \omega \) is the angular speed of the Earth's rotation.

2. **Setting Effective Gravity to Zero**:
To find the angular speed \( \omega_1 \) that makes the effective gravity zero at the equator, we set:
\[ g_{\text{effective}} = 0 \]
Thus, we have:
\[ 0 = g - r \omega_1^2 \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ r \omega_1^2 = g \]

4. **Solving for \( \omega_1 \)**:
Dividing both sides by \( r \):
\[ \omega_1^2 = \frac{g}{r} \]
Taking the square root of both sides, we find:
\[ \omega_1 = \sqrt{\frac{g}{r}} \]

5. **Final Result**:
The final angular speed \( \omega_1 \) required to make the effective acceleration due to gravity equal to zero at the equator is:
\[ \omega_1 = \sqrt{\frac{g}{r}} \]