A star can be considered as spherical ball of hot gas of radius `R`. Inside the star, the density of the gas is `rho_(r)` at radius `r` and mass of the gas within this region is `M_(r)`.
The correct differential equation for variation of mass with respect to radius is (refer to the adjacent figure)

Answer

To derive the correct differential equation for the variation of mass with respect to radius inside a star, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between mass, density, and volume**:
The mass \( M(r) \) of a spherical shell of gas at radius \( r \) can be expressed as:
\[ M(r) = \rho(r) \cdot V \]
where \( V \) is the volume of the sphere of radius \( r \).

2. **Calculate the volume of the sphere**:
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, we can rewrite the mass as:
\[ M(r) = \rho(r) \cdot \frac{4}{3} \pi r^3 \]

3. **Differentiate the mass with respect to radius**:
To find the variation of mass with respect to radius, we differentiate \( M(r) \) with respect to \( r \):
\[ \frac{dM}{dr} = \frac{d}{dr} \left( \rho(r) \cdot \frac{4}{3} \pi r^3 \right) \]

4. **Apply the product rule for differentiation**:
Using the product rule, we have:
\[ \frac{dM}{dr} = \frac{4}{3} \pi \left( \frac{d\rho}{dr} r^3 + \rho(r) \cdot 3r^2 \right) \]

5. **Simplify the expression**:
This can be simplified to:
\[ \frac{dM}{dr} = \frac{4}{3} \pi \left( r^3 \frac{d\rho}{dr} + 3\rho(r) r^2 \right) \]

6. **Final form of the differential equation**:
The differential equation for the variation of mass with respect to radius can be expressed as:
\[ \frac{dM}{dr} = 4 \pi r^2 \rho(r) \]
(assuming the term involving \( \frac{d\rho}{dr} \) is negligible for a uniform density distribution).

### Final Answer:
The correct differential equation for the variation of mass with respect to radius inside the star is:
\[ \frac{dM}{dr} = 4 \pi r^2 \rho(r) \]