A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator ?
(Use `g =(GM)/R^2=9.8 ms^(-2)` and radius of earth, R = 6400 km.]

Options

A. 49 N
B. 49.83 N
C. 48.83 N
D. 49.17 N

Answer

To solve the problem, we need to calculate the weight of a body at the equator given its weight at the North Pole. The weight of the body is affected by the acceleration due to gravity, which varies slightly between the North Pole and the equator due to the Earth's rotation.

### Step-by-Step Solution:

1. **Understanding Weight at the North Pole**:
The weight of the body at the North Pole is given as 49 N. The weight (W) can be expressed as:
\[ W = m \cdot g_p \]
where \( m \) is the mass of the body and \( g_p \) is the acceleration due to gravity at the North Pole.

2. **Finding the Mass of the Body**:
We know that at the North Pole, the acceleration due to gravity \( g_p \) is \( 9.8 \, \text{m/s}^2 \). We can rearrange the weight equation to find the mass:
\[ m = \frac{W}{g_p} = \frac{49 \, \text{N}}{9.8 \, \text{m/s}^2} = 5 \, \text{kg} \]

3. **Calculating Effective Gravity at the Equator**:
The effective acceleration due to gravity at the equator \( g_e \) is given by:
\[ g_e = g - R_e \cdot \omega^2 \]
where:
- \( g = 9.8 \, \text{m/s}^2 \)
- \( R_e = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)
- \( \omega \) is the angular speed of the Earth. The angular speed can be calculated as:
\[ \omega = \frac{2\pi}{T} \]
where \( T = 24 \times 60 \times 60 \, \text{s} \).

4. **Calculating \( \omega \)**:
\[ T = 86400 \, \text{s} \quad \Rightarrow \quad \omega = \frac{2\pi}{86400} \approx 7.272 \times 10^{-5} \, \text{rad/s} \]

5. **Calculating \( R_e \cdot \omega^2 \)**:
\[ R_e \cdot \omega^2 = (6400 \times 10^3) \cdot (7.272 \times 10^{-5})^2 \approx 6400 \times 10^3 \cdot 5.29 \times 10^{-9} \approx 33.8 \, \text{m/s}^2 \]

6. **Finding \( g_e \)**:
\[ g_e = 9.8 - 0.0338 \approx 9.7662 \, \text{m/s}^2 \]

7. **Calculating Weight at the Equator**:
Now we can find the weight of the body at the equator using the mass we calculated:
\[ W_e = m \cdot g_e = 5 \, \text{kg} \cdot 9.7662 \, \text{m/s}^2 \approx 48.83 \, \text{N} \]

### Final Answer:
The weight of the body when shifted to the equator will be approximately **48.83 N**.