Let `omega` be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth `d` below earth's surface at a pole `(dlt ltR)`. the value of `d` is-

Options

A. `(omega^(2)R^(2))/(g)`
B. `(omega^(2)R^(2))/(2g)`
C. `(2omega^(2)R^(2))/(g)`
D. `(omega^(2)R^(2))/(4g)`

Answer

To solve the problem, we need to find the depth \( d \) below the Earth's surface at the poles where the weight of an object is the same as its weight at the equator. We will use the concepts of gravitational acceleration and the effects of rotation.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- At the equator, the weight of an object is affected by the gravitational force \( g \) and the centrifugal force due to the Earth's rotation.
- At the poles, there is no centrifugal force acting on the object since the rotational motion is not present at the poles.

2. **Weight at the Equator**:
- The effective gravitational acceleration at the equator, considering the centrifugal force, is given by:
\[ g_{eq} = g - R \omega^2 \]
where \( R \) is the radius of the Earth and \( \omega \) is the angular velocity of the Earth's rotation.

3. **Weight at Depth \( d \)**:
- At a depth \( d \) below the surface of the Earth, the gravitational acceleration is modified by the depth. The effective gravitational acceleration at this depth is given by:
\[ g_{d} = g \left(1 - \frac{d}{R}\right) \]

4. **Setting the Weights Equal**:
- According to the problem, the weight at the equator is equal to the weight at depth \( d \) at the pole:
\[ g - R \omega^2 = g \left(1 - \frac{d}{R}\right) \]

5. **Rearranging the Equation**:
- Rearranging the above equation gives:
\[ g - R \omega^2 = g - \frac{gd}{R} \]
- Simplifying this leads to:
\[ R \omega^2 = \frac{gd}{R} \]

6. **Solving for Depth \( d \)**:
- Rearranging for \( d \) gives:
\[ d = \frac{R^2 \omega^2}{g} \]

### Final Answer:
The value of \( d \) is:
\[ d = \frac{R^2 \omega^2}{g} \]