The radius of earth is about 6400 km and that of Mars is 3200 km . The mass of the earth is about 10 times the mass of Mars . An object weighs 200 N on te surface of Earth . Its weight on the surface of mars will be .

Options

A. 20 N
B. 8 N
C. 80 N
D. 40 N

Answer

To solve the problem, we need to determine the weight of an object on the surface of Mars given its weight on Earth. We will use the relationship between gravitational acceleration (g) on different planets and the weight of the object.

### Step-by-Step Solution:

1. **Understand the relationship between weight 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.

2. **Weight on Earth:**
We know the weight of the object on Earth (\( W_E \)) is 200 N. Therefore:
\[ W_E = m \cdot g_E \]
where \( g_E \) is the gravitational acceleration on Earth.

3. **Weight on Mars:**
We want to find the weight of the object on Mars (\( W_M \)):
\[ W_M = m \cdot g_M \]
where \( g_M \) is the gravitational acceleration on Mars.

4. **Relate the gravitational accelerations:**
The gravitational acceleration on a planet is given by:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

5. **Given values:**
- Radius of Earth (\( R_E \)) = 6400 km
- Radius of Mars (\( R_M \)) = 3200 km
- Mass of Earth (\( M_E \)) = 10 times the mass of Mars (\( M_M \)), so \( M_E = 10 M_M \).

6. **Calculate the ratio of gravitational accelerations:**
\[ \frac{g_E}{g_M} = \frac{M_E / R_E^2}{M_M / R_M^2} = \frac{10 M_M / (6400^2)}{M_M / (3200^2)} = \frac{10 \cdot 3200^2}{6400^2} \]
Simplifying this:
\[ \frac{g_E}{g_M} = 10 \cdot \left(\frac{3200}{6400}\right)^2 = 10 \cdot \left(\frac{1}{2}\right)^2 = 10 \cdot \frac{1}{4} = \frac{10}{4} = 2.5 \]
Thus, \( g_M = \frac{g_E}{2.5} \).

7. **Substituting back to find \( W_M \):**
From the relationship of weights:
\[ W_M = m \cdot g_M = m \cdot \left(\frac{g_E}{2.5}\right) \]
Since \( W_E = m \cdot g_E \), we can express \( m \) as:
\[ m = \frac{W_E}{g_E} = \frac{200}{g_E} \]
Therefore:
\[ W_M = \frac{200}{g_E} \cdot \left(\frac{g_E}{2.5}\right) = \frac{200}{2.5} = 80 \text{ N} \]

### Final Answer:
The weight of the object on the surface of Mars is **80 N**.