If the density of a planet is constant, then the curve between value of g on its surface and its radius r will be-

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to establish the relationship between the gravitational acceleration \( g \) at the surface of a planet and its radius \( r \), given that the density of the planet is constant.

### Step-by-Step Solution:

1. **Understanding the relationship**: The gravitational acceleration \( g \) at the surface of a planet is given by the formula:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Expressing mass in terms of density**: Since the density \( \rho \) of the planet is constant, we can express the mass \( M \) in terms of density and volume:
\[ M = \rho \cdot V \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, we can write:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Substituting mass back into the equation for \( g \)**: Now, substituting the expression for \( M \) into the equation for \( g \):
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]

4. **Simplifying the equation**: This simplifies to:
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R \]
Here, we can see that \( g \) is directly proportional to \( R \):
\[ g \propto R \]

5. **Interpreting the relationship**: Since \( g \) is directly proportional to \( R \), if we plot \( g \) against \( R \), we will get a straight line that passes through the origin.

### Conclusion:
Thus, the curve between the value of \( g \) on the surface of the planet and its radius \( r \) will be a straight line passing through the origin.

### Final Answer:
The answer is option A: A straight line passing through the origin.

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