Find the duration of the day if the rotation of earth about its axis speeds up in such a way that a person on the equator feels weightlessness.

Options

A. `2pi sqrt((R^(2))/(g^(3)))`
B. `(1)/(2pi) sqrt((R )/(g))`
C. `2pisqrt((R )/(g))`
D. `(1)/(2pi)sqrt((R )/(g^(3)))`

Explanation

To find the duration of the day when the rotation of the Earth speeds up such that a person on the equator feels weightlessness, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Weightlessness at the Equator**:
- A person feels weightlessness when the effective gravitational force acting on them becomes zero. The effective gravitational force \( g' \) at the equator can be expressed as:
\[ g' = g - r \omega^2 \]
- Here, \( g \) is the acceleration due to gravity, \( r \) is the radius of the Earth, and \( \omega \) is the angular speed of the Earth.

2. **Setting Up the Equation for Weightlessness**:
- For a person to feel weightless, we set \( g' = 0 \):
\[ 0 = g - r \omega^2 \]
- Rearranging this gives:
\[ g = r \omega^2 \]

3. **Relating Angular Speed to the Duration of the Day**:
- The angular speed \( \omega \) is related to the duration of the day \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
- Substituting this expression for \( \omega \) into the equation \( g = r \omega^2 \):
\[ g = r \left(\frac{2\pi}{T}\right)^2 \]

4. **Rearranging to Find the Duration of the Day**:
- Rearranging the equation gives:
\[ T^2 = \frac{4\pi^2 r}{g} \]
- Taking the square root of both sides, we find:
\[ T = 2\pi \sqrt{\frac{r}{g}} \]

5. **Final Expression for Duration of the Day**:
- Therefore, the duration of the day when a person at the equator feels weightlessness is:
\[ T = 2\pi \sqrt{\frac{r}{g}} \]