If all objects on the equator of earth feels weightless then the duration of the day will nearly become

Options

A. 6.2 hr
B. 4.4 hr
C. 2.2 hr
D. 1.41 hr

Answer

To solve the problem, we need to determine the duration of a day on Earth if all objects at the equator feel weightless. This situation occurs when the effective acceleration due to gravity (g') becomes zero due to the rotational effects of the Earth.

### Step-by-Step Solution:

1. **Understanding Effective Gravity**:
The effective acceleration due to gravity at the equator can be expressed as:
\[ g' = g - R_e \omega^2 \cos^2(\lambda) \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R_e \) is the radius of the Earth (approximately \( 6400 \, \text{km} \) or \( 6.4 \times 10^6 \, \text{m} \)),
- \( \omega \) is the angular speed of the Earth,
- \( \lambda \) is the latitude (which is \( 0 \) at the equator).

2. **Setting Conditions for Weightlessness**:
For objects to feel weightless, we need:
\[ g' = 0 \]
This leads to:
\[ 0 = g - R_e \omega^2 \]
Rearranging gives:
\[ R_e \omega^2 = g \]

3. **Solving for Angular Speed (\(\omega\))**:
We can solve for \(\omega\):
\[ \omega = \sqrt{\frac{g}{R_e}} \]

4. **Finding the Period of Rotation (T)**:
The period of rotation (T) is related to angular speed by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \(\omega\):
\[ T = \frac{2\pi}{\sqrt{\frac{g}{R_e}}} \]

5. **Simplifying the Expression**:
This can be rewritten as:
\[ T = 2\pi \sqrt{\frac{R_e}{g}} \]

6. **Calculating the Time Period**:
Plugging in the values for \( R_e \) and \( g \):
\[ R_e \approx 6.4 \times 10^6 \, \text{m}, \quad g \approx 9.81 \, \text{m/s}^2 \]
\[ T = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.81}} \approx 2\pi \sqrt{651,000} \approx 2\pi \times 806.5 \approx 5065 \, \text{s} \]

7. **Converting Seconds to Hours**:
To convert seconds into hours:
\[ \text{Hours} = \frac{5065 \, \text{s}}{3600 \, \text{s/hour}} \approx 1.41 \, \text{hours} \]

### Final Answer:
If all objects on the equator of Earth feel weightless, the duration of the day will nearly become **1.41 hours**.