The time period of earth's rotation should be `x pi sqrt((R_(e))/(g))` if the weight of a body at equator become 3/4 times of its actual weight. Find the vlaue of x.

Answer

To solve the problem, we need to find the value of \( x \) in the equation for the time period of Earth's rotation given that the weight of a body at the equator becomes \( \frac{3}{4} \) times its actual weight.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We know that the effective weight of a body at the equator is given as \( \frac{3}{4} \) of its actual weight. The actual weight \( W \) of a body is given by \( W = mg \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

2. **Setting Up the Effective Weight:**
The effective weight at the equator can be expressed as:
\[ W' = \frac{3}{4} W = \frac{3}{4} mg \]
The effective weight can also be expressed in terms of the effective gravity \( g_{\text{effective}} \):
\[ W' = m g_{\text{effective}} \implies g_{\text{effective}} = \frac{3}{4} g \]

3. **Relating Effective Gravity to Centripetal Acceleration:**
The effective gravity at the equator is given by:
\[ g_{\text{effective}} = g - R_e \omega^2 \]
where \( R_e \) is the radius of the Earth and \( \omega \) is the angular velocity of the Earth's rotation.

4. **Substituting for Effective Gravity:**
From the previous step, we can equate the two expressions for \( g_{\text{effective}} \):
\[ \frac{3}{4} g = g - R_e \omega^2 \]
Rearranging gives:
\[ R_e \omega^2 = g - \frac{3}{4} g = \frac{1}{4} g \]

5. **Finding Angular Velocity:**
Now, we can express \( \omega^2 \):
\[ \omega^2 = \frac{g}{4 R_e} \]

6. **Relating Angular Velocity to Time Period:**
The angular velocity \( \omega \) is also related to the time period \( T \) of rotation:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into our equation gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{g}{4 R_e} \]
Rearranging for \( T \):
\[ T^2 = \frac{4\pi^2 R_e}{g} \]
Taking the square root:
\[ T = 2\pi \sqrt{\frac{R_e}{g}} \]

7. **Expressing Time Period in Terms of \( x \):**
We are given that:
\[ T = x \pi \sqrt{\frac{R_e}{g}} \]
Setting the two expressions for \( T \) equal gives:
\[ 2\pi \sqrt{\frac{R_e}{g}} = x \pi \sqrt{\frac{R_e}{g}} \]

8. **Solving for \( x \):**
Dividing both sides by \( \pi \sqrt{\frac{R_e}{g}} \) (assuming it's not zero):
\[ 2 = x \]

Thus, the value of \( x \) is:

\[ \boxed{4} \]