The radius of earth is about `6400Km` and that of mars is about `3200 km` The mass of the earth is about `10`times the mass of mars. An object weight `200N` on earth 's surface, then its weight on the surface of mars will be:

Options

A. `6N`
B. `20N`
C. `40N`
D. `80N`

Explanation

To find the weight of an object on the surface of Mars given its weight on Earth, we can follow these steps:

### Step 1: Understand the relationship between weight, mass, and gravitational acceleration
The weight \( W \) of an object is given by the formula:
\[ W = m \cdot g \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

### Step 2: Define the known values
- Weight of the object on Earth, \( W_E = 200 \, \text{N} \)
- Radius of Earth, \( R_E = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)
- Radius of Mars, \( R_M = 3200 \, \text{km} = 3200 \times 10^3 \, \text{m} \)
- Mass of Earth, \( M_E \)
- Mass of Mars, \( M_M = \frac{M_E}{10} \) (since the mass of Earth is 10 times that of Mars)

### Step 3: Calculate the acceleration due to gravity on Earth
Using the formula for gravitational acceleration:
\[ g_E = \frac{G \cdot M_E}{R_E^2} \]
where \( G \) is the gravitational constant.

### Step 4: Calculate the acceleration due to gravity on Mars
Using the same formula for Mars:
\[ g_M = \frac{G \cdot M_M}{R_M^2} \]
Substituting \( M_M \):
\[ g_M = \frac{G \cdot \frac{M_E}{10}}{R_M^2} \]

### Step 5: Relate the gravitational accelerations
Now, we can express \( g_M \) in terms of \( g_E \):
\[ g_M = \frac{G \cdot \frac{M_E}{10}}{R_M^2} = \frac{1}{10} \cdot \frac{G \cdot M_E}{R_M^2} \]
We can also express \( g_E \) in terms of \( R_E \):
\[ g_E = \frac{G \cdot M_E}{R_E^2} \]
Now, we can find the ratio of \( g_M \) to \( g_E \):
\[ \frac{g_M}{g_E} = \frac{\frac{1}{10} \cdot \frac{G \cdot M_E}{R_M^2}}{\frac{G \cdot M_E}{R_E^2}} = \frac{R_E^2}{10 \cdot R_M^2} \]

### Step 6: Substitute the values of \( R_E \) and \( R_M \)
Substituting the values:
\[ \frac{g_M}{g_E} = \frac{(6400 \times 10^3)^2}{10 \cdot (3200 \times 10^3)^2} \]
Calculating this gives:
\[ \frac{g_M}{g_E} = \frac{6400^2}{10 \cdot 3200^2} = \frac{6400^2}{3200^2 \cdot 10} = \frac{4}{10} = \frac{2}{5} \]
Thus, we find:
\[ g_M = \frac{2}{5} g_E \]

### Step 7: Calculate the weight on Mars
Using the relationship between weight and gravitational acceleration:
\[ W_M = m \cdot g_M \]
From the weight on Earth:
\[ W_E = m \cdot g_E \Rightarrow m = \frac{W_E}{g_E} = \frac{200}{g_E} \]
Now substituting \( g_M \):
\[ W_M = \frac{200}{g_E} \cdot g_M = \frac{200}{g_E} \cdot \left(\frac{2}{5} g_E\right) = 200 \cdot \frac{2}{5} = 80 \, \text{N} \]

### Final Answer
The weight of the object on the surface of Mars is \( 80 \, \text{N} \).

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