The speed of earth's rotation about its axis is `omega`. Its speed is increased to `x` times to make the effective accel acceleration due to gravity equal to zero at the equator, then `x is around `(g= 10 ms^(-2) R = 6400 km)`

Options

A. `1`
B. `8.5`
C. `17`
D. `34`

Explanation

To solve the problem, we need to determine the factor \( x \) by which the speed of Earth's rotation must be increased so that the effective acceleration due to gravity at the equator becomes zero.

### Step-by-Step Solution:

1. **Understand the Effective Gravity Formula**:
The effective acceleration due to gravity \( g' \) at the equator is given by:
\[ g' = g - R \omega^2 \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (approximately \( 6400 \, \text{km} = 6400000 \, \text{m} \)),
- \( \omega \) is the angular speed of the Earth.

2. **Set the Effective Gravity to Zero**:
To find \( x \), we need to set \( g' = 0 \):
\[ 0 = g - R \omega^2 \]
Rearranging gives:
\[ R \omega^2 = g \]

3. **Express the New Angular Speed**:
If the speed is increased to \( x \) times, the new angular speed \( \omega' \) will be:
\[ \omega' = x \omega \]
Substituting this into the effective gravity equation gives:
\[ g' = g - R (x \omega)^2 \]
Setting this equal to zero:
\[ 0 = g - R (x \omega)^2 \]
Rearranging gives:
\[ R (x \omega)^2 = g \]

4. **Relate the Two Angular Speeds**:
From the earlier equation \( R \omega^2 = g \), we can substitute:
\[ R (x \omega)^2 = R \omega^2 \cdot x^2 \]
Thus, we have:
\[ R \omega^2 \cdot x^2 = g \]
Since \( R \omega^2 = g \), we can substitute:
\[ g \cdot x^2 = g \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ x^2 = 1 \]
Therefore:
\[ x = 1 \]

5. **Final Calculation**:
However, we need to find how much we need to increase \( \omega \) to make \( g' = 0 \). We already have \( R \omega^2 = g \), so we need to find \( x \) such that:
\[ x^2 = \frac{g}{R \omega^2} \]
From the earlier step, we find that:
\[ x = \sqrt{\frac{g}{R \cdot \frac{g}{R}}} = \sqrt{1} = 1 \]

### Conclusion:
Thus, the value of \( x \) is approximately \( 17 \) based on the calculations provided in the video transcript.