A body weighs 72 N on surface of earth feel 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 the day when a body weighing 72 N on the surface of the Earth feels weightless. This situation occurs when the effective gravitational force acting on the body becomes zero due to the centrifugal force caused by the Earth's rotation.

### Step-by-Step Solution:

1. **Understanding the Forces**:
When a body feels weightless, the effective gravitational force acting on it is zero. The effective gravitational force (G') can be expressed as:
\[ G' = G - R_e \omega^2 \]
where \( G \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), \( R_e \) is the radius of the Earth, and \( \omega \) is the angular velocity of the Earth.

2. **Setting Up the Equation**:
For the body to feel weightless:
\[ G' = 0 \implies G = R_e \omega^2 \]

3. **Finding Angular Velocity**:
Rearranging the equation gives:
\[ \omega^2 = \frac{G}{R_e} \]
Taking the square root:
\[ \omega = \sqrt{\frac{G}{R_e}} \]

4. **Substituting Values**:
We know:
- \( G \approx 9.81 \, \text{m/s}^2 \)
- The average radius of the Earth \( R_e \approx 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Plugging in the values:
\[ \omega = \sqrt{\frac{9.81}{6400 \times 10^3}} \approx \sqrt{\frac{9.81}{6400000}} \approx \sqrt{1.53 \times 10^{-6}} \approx 0.001236 \, \text{rad/s} \]

5. **Calculating the Duration of the Day**:
The period \( T \) of rotation (the duration of the day) is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{0.001236} \approx 5085.7 \, \text{s} \]

6. **Converting to Hours**:
To convert seconds into hours:
\[ T \approx \frac{5085.7}{3600} \approx 1.41 \, \text{hours} \]

### Final Answer:
The duration of the day will nearly become **1.41 hours**.

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