The speed of earth's rotation about its axis is `omega`. Its speed is increased to `x` times to make the effective 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 the Earth's rotation must be increased to make the effective acceleration due to gravity at the equator equal to zero.

### Step-by-Step Solution:

1. **Understanding Effective Gravity**:
The effective acceleration due to gravity \( g_{\text{effective}} \) at the equator is given by:
\[ g_{\text{effective}} = g - \omega^2 R \]
where \( g \) is the gravitational acceleration (approximately \( 10 \, \text{m/s}^2 \)), \( \omega \) is the angular speed of the Earth, and \( R \) is the radius of the Earth (approximately \( 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)).

2. **Setting Effective Gravity to Zero**:
We want to find \( x \) such that:
\[ g - \omega_2^2 R = 0 \]
where \( \omega_2 = x \omega \). Rearranging gives:
\[ g = \omega_2^2 R \]

3. **Substituting \( \omega_2 \)**:
Substitute \( \omega_2 \) into the equation:
\[ g = (x \omega)^2 R \]
This simplifies to:
\[ g = x^2 \omega^2 R \]

4. **Finding the Original Angular Speed \( \omega \)**:
The original angular speed \( \omega \) can be calculated using the period of rotation of the Earth (24 hours):
\[ \omega = \frac{2\pi}{T_1} = \frac{2\pi}{24 \times 3600} \, \text{s}^{-1} \]

5. **Substituting \( \omega \)**:
Now, substituting \( \omega \) back into the equation:
\[ g = x^2 \left(\frac{2\pi}{24 \times 3600}\right)^2 R \]

6. **Solving for \( x^2 \)**:
Rearranging gives:
\[ x^2 = \frac{g}{R \left(\frac{2\pi}{24 \times 3600}\right)^2} \]

7. **Calculating \( x \)**:
Plugging in the values:
- \( g = 10 \, \text{m/s}^2 \)
- \( R = 6.4 \times 10^6 \, \text{m} \)
- \( T_1 = 24 \times 3600 \, \text{s} \)

Calculate \( \omega \):
\[ \omega = \frac{2\pi}{86400} \approx 7.272 \times 10^{-5} \, \text{rad/s} \]

Now calculate \( \omega^2 R \):
\[ \omega^2 R = (7.272 \times 10^{-5})^2 \times 6.4 \times 10^6 \approx 0.0032 \, \text{m/s}^2 \]

Therefore:
\[ x^2 = \frac{10}{0.0032} \approx 3125 \]
Thus:
\[ x \approx \sqrt{3125} \approx 56 \]

### Final Answer:
The value of \( x \) is approximately \( 56 \).