A planet has double the mass and double the average density of the earth. If the weight of an object on the earth is 100 N, its weight on the planet will be

Options

A. 50 N
B. 100 N
C. 200 N
D. 400 N

Explanation

To solve the problem, we need to determine the weight of an object on a planet that has double the mass and double the average density of the Earth. We know the weight of the object on Earth is 100 N.

### Step-by-Step Solution:

1. **Identify the Mass and Density Relationships**:
- Let the mass of the Earth be \( M_E \) and the density of the Earth be \( \rho_E \).
- The mass of the planet \( M_P \) is given as \( 2M_E \) (double the mass of Earth).
- The density of the planet \( \rho_P \) is given as \( 2\rho_E \) (double the density of Earth).

2. **Relate Mass, Density, and Volume**:
- The mass of a spherical object can be expressed as:
\[ M = \rho \times V \]
- For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, we can express the mass of the Earth and the planet as:
\[ M_E = \rho_E \times \frac{4}{3} \pi r_E^3 \]
\[ M_P = \rho_P \times \frac{4}{3} \pi r_P^3 \]

3. **Set Up the Mass Ratio**:
- From the mass relationships, we have:
\[ \frac{M_P}{M_E} = \frac{\rho_P \times \frac{4}{3} \pi r_P^3}{\rho_E \times \frac{4}{3} \pi r_E^3} \]
- Simplifying this gives:
\[ \frac{M_P}{M_E} = \frac{\rho_P}{\rho_E} \times \left(\frac{r_P}{r_E}\right)^3 \]
- Substituting the known values:
\[ 2 = 2 \times \left(\frac{r_P}{r_E}\right)^3 \]
- This simplifies to:
\[ 1 = \left(\frac{r_P}{r_E}\right)^3 \]
- Therefore, we find:
\[ r_P = r_E \]
- This means the radius of the planet is equal to the radius of the Earth.

4. **Calculate the Weight on the Planet**:
- The weight of an object is given by:
\[ W = m \cdot g \]
- The acceleration due to gravity \( g \) on the surface of a planet is given by:
\[ g = \frac{G \cdot M}{r^2} \]
- For the Earth:
\[ g_E = \frac{G \cdot M_E}{r_E^2} \]
- For the planet:
\[ g_P = \frac{G \cdot M_P}{r_P^2} \]
- Since \( M_P = 2M_E \) and \( r_P = r_E \):
\[ g_P = \frac{G \cdot 2M_E}{r_E^2} = 2 \cdot \frac{G \cdot M_E}{r_E^2} = 2g_E \]

5. **Weight Ratio**:
- The weight of the object on the planet \( W_P \) can be expressed as:
\[ W_P = m \cdot g_P = m \cdot (2g_E) \]
- Since the weight on Earth is \( W_E = m \cdot g_E = 100 \, \text{N} \):
\[ W_P = 2 \cdot W_E = 2 \cdot 100 \, \text{N} = 200 \, \text{N} \]

### Final Answer:
The weight of the object on the planet will be **200 N**.