What should be the angular velocity of earth about its own axis so that the weight of the body at the equator would become `3/4` th of its present value ?

Options

A. `1/400 rads^(-1)`
B. `1/800 rad s^(-1)`
C. `1/1600 rad s^(-1)`
D. `1/3200 rad s^(-1)`

Answer

To find the angular velocity of the Earth about its own axis such that the weight of a body at the equator becomes \( \frac{3}{4} \) of its present value, we can follow these steps:

### Step 1: Understand the forces acting on the body
At the equator, the weight of the body is given by \( mg \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity. When the Earth rotates, there is a centrifugal force acting outward, which can be expressed as \( m \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the radius of the Earth.

### Step 2: Set up the equation
The effective weight \( W' \) of the body at the equator when the centrifugal force is considered can be expressed as:
\[ W' = mg - m\omega^2 r \]
We want this effective weight to be \( \frac{3}{4} mg \):
\[ mg - m\omega^2 r = \frac{3}{4} mg \]

### Step 3: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g - \omega^2 r = \frac{3}{4} g \]
Now, rearranging gives:
\[ g - \frac{3}{4} g = \omega^2 r \]
\[ \frac{1}{4} g = \omega^2 r \]

### Step 4: Solve for \( \omega^2 \)
Now, we can express \( \omega^2 \):
\[ \omega^2 = \frac{g}{4r} \]

### Step 5: Solve for \( \omega \)
Taking the square root gives us:
\[ \omega = \sqrt{\frac{g}{4r}} = \frac{1}{2} \sqrt{\frac{g}{r}} \]

### Step 6: Substitute values
Now, we substitute the known values:
- \( g \approx 10 \, \text{m/s}^2 \)
- \( r \approx 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Calculating \( \omega \):
\[ \omega = \frac{1}{2} \sqrt{\frac{10}{6400 \times 10^3}} \]

### Step 7: Simplify the expression
Calculating \( \frac{10}{6400 \times 10^3} \):
\[ \frac{10}{6400 \times 10^3} = \frac{10}{6.4 \times 10^6} = \frac{1}{640000} = 1.5625 \times 10^{-6} \]
Now taking the square root:
\[ \sqrt{1.5625 \times 10^{-6}} \approx 0.00125 \]
Thus,
\[ \omega \approx \frac{1}{2} \times 0.00125 = 0.000625 \, \text{radians/second} \]

### Final Result
The angular velocity \( \omega \) required for the weight of a body at the equator to become \( \frac{3}{4} \) of its present value is approximately \( 0.000625 \, \text{radians/second} \).

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