The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to
[Take g `= 10 m//s^(2)` for the acceleration due to gravity if the earth were at rest and radius of earth equal to 6400 km.]

Options

A. `1.25xx10^(-3)` rad/s
B. `2.50xx10^(-3)` rad/s
C. `3.75xx10^(-3)` rad/s
D. `5.0xx10^(-3)` rad/s

Answer

To solve the problem, we need to find the imaginary angular velocity (ω) of the Earth for which the effective acceleration due to gravity at the equator becomes zero.

### Step-by-Step Solution:

1. **Understanding the Effective Acceleration Due to Gravity**:
The effective acceleration due to gravity (g') at the equator can be expressed as:
\[ g' = g - \omega^2 r \]
where:
- \( g \) = acceleration due to gravity (10 m/s²)
- \( \omega \) = angular velocity (in rad/s)
- \( r \) = radius of the Earth (in meters)

2. **Setting the Effective Gravity to Zero**:
According to the problem, we want \( g' = 0 \):
\[ 0 = g - \omega^2 r \]
Rearranging this gives:
\[ g = \omega^2 r \]

3. **Solving for Angular Velocity**:
We can solve for \( \omega \):
\[ \omega^2 = \frac{g}{r} \]
Taking the square root of both sides:
\[ \omega = \sqrt{\frac{g}{r}} \]

4. **Substituting the Values**:
Now we substitute the values for \( g \) and \( r \):
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m} \)

Therefore:
\[ \omega = \sqrt{\frac{10}{6.4 \times 10^6}} \]

5. **Calculating the Value**:
First, calculate the fraction:
\[ \frac{10}{6.4 \times 10^6} = 1.5625 \times 10^{-6} \]
Now take the square root:
\[ \omega = \sqrt{1.5625 \times 10^{-6}} = 1.25 \times 10^{-3} \, \text{rad/s} \]

6. **Final Answer**:
Thus, the imaginary angular velocity of the Earth for which the effective acceleration due to gravity at the equator is zero is:
\[ \omega = 1.25 \times 10^{-3} \, \text{rad/s} \]