The reading of a spring balance corresponds to 100 N while situated at the north pole and a body is kept on it. The weight record on the same scale if it is shifted to the equator, is (take, `g = 19 " ms"^(2-)` and radius of the earth, `R = 6.4 xx 10^(6)` m)

Options

A. 99.66 N
B. 110 N
C. 97.66 N
D. 106 N

Explanation

To solve the problem, we need to determine the weight of a body when it is moved from the North Pole to the Equator using the information provided.

### Step-by-Step Solution:

1. **Understanding the Weight at the North Pole**:
The weight of the body at the North Pole is given as 100 N. The weight (W) is related to mass (m) and acceleration due to gravity (g) by the equation:
\[ W = m \cdot g \]
Here, at the North Pole, \( g = 19 \, \text{m/s}^2 \).

2. **Calculating the Mass of the Body**:
Rearranging the weight equation to find mass:
\[ m = \frac{W}{g} = \frac{100 \, \text{N}}{19 \, \text{m/s}^2} \approx 5.26 \, \text{kg} \]

3. **Finding the Angular Speed of the Earth**:
The angular speed (\( \omega \)) of the Earth can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of rotation (24 hours = 86400 seconds):
\[ \omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \, \text{rad/s} \]

4. **Calculating the Apparent Weight at the Equator**:
The apparent weight (\( W' \)) at the Equator is given by:
\[ W' = m \cdot g' = m \cdot (g - \omega^2 R) \]
where \( R \) is the radius of the Earth (given as \( 6.4 \times 10^6 \, \text{m} \)). First, we need to calculate \( \omega^2 R \):
\[ \omega^2 = (7.27 \times 10^{-5})^2 \approx 5.29 \times 10^{-9} \, \text{rad}^2/\text{s}^2 \]
\[ \omega^2 R = 5.29 \times 10^{-9} \cdot 6.4 \times 10^6 \approx 0.0338 \, \text{m/s}^2 \]

5. **Calculating the Effective Gravity at the Equator**:
Now, we can find the effective gravity at the Equator:
\[ g' = g - \omega^2 R = 19 \, \text{m/s}^2 - 0.0338 \, \text{m/s}^2 \approx 18.9662 \, \text{m/s}^2 \]

6. **Calculating the Apparent Weight**:
Finally, substituting back to find the apparent weight at the Equator:
\[ W' = m \cdot g' = 5.26 \, \text{kg} \cdot 18.9662 \, \text{m/s}^2 \approx 99.76 \, \text{N} \]

### Conclusion:
The reading on the spring balance when the body is moved to the Equator is approximately **99.76 N**.

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