When a body is taken poles to equator on the earth, its weight

Options

A. increases
B. decreases
C. remains same
D. increases at south pole and decreases at north pole

Answer

To solve the question of how the weight of a body changes when it is taken from the poles to the equator on Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Weight**:
- Weight (W) of an object is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

2. **Recognize that Mass Remains Constant**:
- When moving a body from the poles to the equator, the mass \( m \) of the body does not change.

3. **Understand the Variation of 'g'**:
- The acceleration due to gravity \( g \) varies with latitude. It can be expressed as:
\[ g = g_0 (1 - \frac{r \omega^2 \cos^2 \theta}{g_0}) \]
where:
- \( g_0 \) is the standard acceleration due to gravity,
- \( r \) is the radius of the Earth,
- \( \omega \) is the angular velocity of the Earth,
- \( \theta \) is the latitude.

4. **Evaluate 'g' at the Poles and the Equator**:
- At the poles (\( \theta = 90^\circ \)), \( \cos(90^\circ) = 0 \), so:
\[ g_{\text{pole}} = g_0 \]
- At the equator (\( \theta = 0^\circ \)), \( \cos(0^\circ) = 1 \), so:
\[ g_{\text{equator}} = g_0 (1 - \frac{r \omega^2}{g_0}) \]
- Since \( r \omega^2 \) is a positive quantity, \( g_{\text{equator}} < g_{\text{pole}} \).

5. **Conclusion on Weight Change**:
- Since the weight \( W \) is directly proportional to \( g \), and \( g \) decreases from the poles to the equator, the weight of the body will also decrease.
- Therefore, the weight at the equator is less than the weight at the poles:
\[ W_{\text{equator}} < W_{\text{pole}} \]

### Final Answer:
When a body is taken from the poles to the equator, its weight decreases.

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