Suppose the earth increases its speed of rotation . At what new time period will the weightof a body on the equator becomes zero? Take `g=10 m/s^2` and radius of the earth `R=6400 km`.

Explanation

To solve the problem, we need to find the new time period of the Earth's rotation at which the weight of a body at the equator becomes zero. We will use the following steps:

### Step-by-Step Solution:

1. **Understanding Weightlessness Condition**:
The weight of a body (W) is given by the equation:
\[ W = mg \]
For the weight to become zero, the effective acceleration due to gravity (g') must be zero. The effective gravity at the equator can be expressed as:
\[ g' = g - \omega^2 R \]
where \( g \) is the acceleration due to gravity, \( \omega \) is the angular velocity of the Earth, and \( R \) is the radius of the Earth.

2. **Setting Up the Equation**:
To find the condition for weightlessness, we set \( g' = 0 \):
\[ 0 = g - \omega^2 R \]
Rearranging gives:
\[ \omega^2 R = g \]
Thus,
\[ \omega = \sqrt{\frac{g}{R}} \]

3. **Substituting Known Values**:
We are given \( g = 10 \, \text{m/s}^2 \) and \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \). Now we can substitute these values into the equation:
\[ \omega = \sqrt{\frac{10}{6400 \times 10^3}} \]

4. **Calculating Angular Velocity**:
Performing the calculations:
\[ \omega = \sqrt{\frac{10}{6400000}} = \sqrt{\frac{1}{640000}} = \frac{1}{800} \, \text{rad/s} \]

5. **Finding the Time Period**:
The time period \( T \) is related to angular velocity \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = \frac{2\pi}{\frac{1}{800}} = 1600\pi \, \text{s} \]

6. **Converting to Hours**:
To convert seconds to hours, we divide by 3600:
\[ T_{\text{hours}} = \frac{1600\pi}{3600} = \frac{1600\pi}{3600} \approx 1.4 \, \text{hours} \]

### Final Answer:
The new time period at which the weight of a body on the equator becomes zero is approximately \( 1.4 \, \text{hours} \).