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. 194.32
B. 194.66
C. 195.33
D. 195.2

Explanation

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:
\[ W_p = 196 \, \text{N} \]
Using the formula for weight:
\[ W = m \cdot g \]
where \( g \) is the acceleration due to gravity. At the pole, \( g \) is given as \( 10 \, \text{m/s}^2 \).

### Step 2: Calculate the Mass of the Object
From the weight at the pole, we can calculate 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 Apparent Weight at the Equator
At the equator, the object experiences a centrifugal force due to the Earth's rotation, which affects its apparent weight. The apparent weight \( W_e \) can be calculated using:
\[ W_e = m \cdot g - F_c \]
where \( F_c \) is the centrifugal force given by:
\[ F_c = m \cdot \omega^2 \cdot r \]
Here, \( \omega \) is the angular velocity of the Earth, and \( r \) is the radius of the Earth.

### Step 4: Calculate Angular Velocity \( \omega \)
The Earth completes one rotation in 24 hours. To convert this to seconds:
\[ T = 24 \times 3600 \, \text{s} = 86400 \, \text{s} \]
The angular velocity \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{86400} \, \text{rad/s} \]

### Step 5: Calculate the Radius in Meters
The radius of the Earth is given as \( 6400 \, \text{km} \), which we convert to meters:
\[ r = 6400 \times 10^3 \, \text{m} = 6400000 \, \text{m} \]

### Step 6: Calculate Centrifugal Force \( F_c \)
Substituting \( m \), \( \omega \), and \( r \) into the centrifugal force formula:
\[ F_c = m \cdot \omega^2 \cdot r \]
Calculating \( \omega^2 \):
\[ \omega^2 = \left(\frac{2\pi}{86400}\right)^2 \]
Now substituting the values:
\[ F_c = 19.6 \cdot \left(\frac{2\pi}{86400}\right)^2 \cdot 6400000 \]

### Step 7: Calculate the Apparent Weight at the Equator
Now we can substitute \( F_c \) back into the equation for apparent weight:
\[ W_e = m \cdot g - F_c \]
Substituting the known values:
\[ W_e = 19.6 \cdot 10 - F_c \]

### Step 8: Final Calculation
Perform the calculations to find \( W_e \). After calculating \( F_c \) and substituting it into the equation, we will find the apparent weight at the equator.

### Conclusion
After performing all calculations, we find that the weight at the equator is approximately:
\[ W_e \approx 195.33 \, \text{N} \]