If weight of an object at pole is 196 N then weight at equator is [`g = 10 m/s^2` , radius of earth = 6400 Km]

Options

A. 194.66 N
B. 195.32 N
C. 195.66 N
D. 194.32 N

Answer

To solve the problem of finding the weight of an object at the equator given its weight at the pole, we can follow these steps:

### Step 1: Understand the Weight at the Pole
The weight of the object at the pole is given as \( W_p = 196 \, \text{N} \). This weight is the gravitational force acting on the object, which can be expressed as:
\[ W_p = mg \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the Mass of the Object
Using the weight at the pole, we can find the mass \( m \):
\[ m = \frac{W_p}{g} = \frac{196 \, \text{N}}{10 \, \text{m/s}^2} = 19.6 \, \text{kg} \]

### Step 3: Determine the Angular Velocity of the Earth
The angular velocity \( \omega \) of the Earth can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of rotation of the Earth. Since the Earth takes 24 hours to complete one rotation, we convert this into seconds:
\[ T = 24 \, \text{hours} = 24 \times 60 \times 60 \, \text{seconds} = 86400 \, \text{seconds} \]
Now substituting \( T \) into the angular velocity formula:
\[ \omega = \frac{2\pi}{86400} \, \text{rad/s} \approx 7.272 \times 10^{-5} \, \text{rad/s} \]

### Step 4: Calculate the Centrifugal Force at the Equator
The centrifugal force \( F_c \) acting on the object at the equator is given by:
\[ F_c = m \omega^2 R \]
where \( R \) is the radius of the Earth. Given \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \), we can substitute the values:
\[ F_c = 19.6 \, \text{kg} \times (7.272 \times 10^{-5} \, \text{rad/s})^2 \times (6400 \times 10^3 \, \text{m}) \]

### Step 5: Calculate the Apparent Weight at the Equator
The apparent weight \( W_e \) at the equator can be expressed as:
\[ W_e = mg - F_c \]
Substituting \( mg = 196 \, \text{N} \) and calculating \( F_c \):
1. Calculate \( \omega^2 \):
\[ \omega^2 \approx (7.272 \times 10^{-5})^2 \approx 5.29 \times 10^{-9} \, \text{rad}^2/\text{s}^2 \]
2. Calculate \( F_c \):
\[ F_c = 19.6 \times 5.29 \times 10^{-9} \times 6400 \times 10^3 \approx 0.66 \, \text{N} \]
3. Now substitute to find \( W_e \):
\[ W_e = 196 \, \text{N} - 0.66 \, \text{N} = 195.34 \, \text{N} \]

### Final Answer
Thus, the weight of the object at the equator is approximately:
\[ W_e \approx 195.34 \, \text{N} \]