The acceleration due to gravity at the poles is `10ms^(-2)` and equitorial radius is 6400km for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to 75% is

Options

A. `1/1600 "rads"^(-1)`
B. `1/800 "rads"^(-1)`
C. `1/400 "rads"^(-1)`
D. `1/200 "rads"^(-1)`

Answer

To solve the problem, we need to find the angular velocity of the Earth such that the weight of a body at the equator is reduced to 75% of its weight at the poles. Here’s how we can approach this step-by-step:

### Step 1: Understand the relationship between gravity at the poles and at the equator.
The acceleration due to gravity at the poles (GP) is given as \(10 \, \text{m/s}^2\). The acceleration due to gravity at the equator (GE) can be expressed as:
\[ GE = \frac{3}{4} GP \]
This means:
\[ GE = \frac{3}{4} \times 10 \, \text{m/s}^2 = 7.5 \, \text{m/s}^2 \]

### Step 2: Use the formula for gravitational acceleration at the equator.
The gravitational acceleration at the equator can also be expressed as:
\[ GE = GP - R \omega^2 \]
where \(R\) is the radius of the Earth and \(\omega\) is the angular velocity.

### Step 3: Substitute the known values into the equation.
We know:
- \(GP = 10 \, \text{m/s}^2\)
- \(GE = 7.5 \, \text{m/s}^2\)
- \(R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m}\)

Now substituting these values into the equation:
\[ 7.5 = 10 - (6.4 \times 10^6) \omega^2 \]

### Step 4: Rearrange the equation to solve for \(\omega^2\).
Rearranging gives:
\[ (6.4 \times 10^6) \omega^2 = 10 - 7.5 \]
\[ (6.4 \times 10^6) \omega^2 = 2.5 \]

### Step 5: Solve for \(\omega^2\).
Now, divide both sides by \(6.4 \times 10^6\):
\[ \omega^2 = \frac{2.5}{6.4 \times 10^6} \]

### Step 6: Calculate \(\omega\).
Now we can calculate \(\omega\):
\[ \omega = \sqrt{\frac{2.5}{6.4 \times 10^6}} \]
Calculating this gives:
\[ \omega = \sqrt{\frac{2.5}{6400000}} = \sqrt{3.90625 \times 10^{-7}} \approx 1.975 \times 10^{-4} \, \text{rad/s} \]

### Final Answer:
Thus, the angular velocity of rotation of the Earth about its axis so that the weight of a body at the equator reduces to 75% is approximately:
\[ \omega \approx 1.975 \times 10^{-4} \, \text{rad/s} \]