An insulating solid sphere of radius R has a uniformly positive charge density `rho`. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero.
Statement-1: When a charge 'q' is taken from the centre to the surface of the sphere, its potential energy changes by `(qrho)/(3 in_(0))`
Statement-2 : The electric field at a distance `r (r lt R)` from the centre of the the sphere is `(rho r)/(3in_(0))`

Options

A. Both Statement-1 & Statement-2 are correct & Statement-2 is the correct explanation of Statement-1
B. Both Statement-1 & Statement-2 are correct & Statement-2 is not the correct explanation of Statement-1
C. Statement - 1 is correct but Statement -2 is incorrect
D. Statement-2 is correct but Statement-1 is incorrect

Explanation

To solve the problem, we will analyze both statements provided and derive the necessary equations step by step.

### Step 1: Electric Field Inside the Sphere
For a uniformly charged insulating solid sphere of radius \( R \) with a uniform charge density \( \rho \), we can use Gauss's Law to find the electric field inside the sphere at a distance \( r \) from the center (where \( r < R \)).

Using Gauss's Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Where \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface.

The charge enclosed within a radius \( r \) is given by:
\[ Q_{\text{enc}} = \rho \cdot \text{Volume} = \rho \cdot \frac{4}{3} \pi r^3 \]

The area \( A \) of the Gaussian surface (a sphere of radius \( r \)) is:
\[ A = 4 \pi r^2 \]

Substituting into Gauss's Law:
\[ E \cdot 4 \pi r^2 = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_0} \]

Solving for \( E \):
\[ E = \frac{\rho r}{3 \epsilon_0} \]

### Step 2: Change in Potential Energy
To find the change in potential energy when moving a charge \( q \) from the center of the sphere to the surface, we need to calculate the potential difference \( V \) between these two points.

The electric potential \( V \) at a distance \( r \) from the center is given by:
\[ V = -\int E \, dr \]

The potential at the center (where \( r = 0 \)) is:
\[ V(0) = 0 \quad \text{(arbitrarily chosen as reference)} \]

The potential at the surface \( V(R) \) can be calculated as:
\[ V(R) = -\int_0^R \frac{\rho r}{3 \epsilon_0} \, dr \]

Calculating the integral:
\[ V(R) = -\left[ \frac{\rho}{3 \epsilon_0} \cdot \frac{r^2}{2} \right]_0^R = -\frac{\rho R^2}{6 \epsilon_0} \]

The change in potential energy \( \Delta U \) when moving charge \( q \) from the center to the surface is:
\[ \Delta U = q \cdot (V(R) - V(0)) = q \cdot \left(-\frac{\rho R^2}{6 \epsilon_0} - 0\right) = -\frac{q \rho R^2}{6 \epsilon_0} \]

### Step 3: Analyzing the Statements
1. **Statement 1**: The change in potential energy when a charge \( q \) is taken from the center to the surface is given as \( \frac{q \rho}{3 \epsilon_0} \). Our calculation shows it is \( -\frac{q \rho R^2}{6 \epsilon_0} \). Therefore, **Statement 1 is false**.

2. **Statement 2**: The electric field at a distance \( r < R \) is given as \( \frac{\rho r}{3 \epsilon_0} \). Our calculation confirms this is true. Therefore, **Statement 2 is true**.

### Conclusion
- **Statement 1**: False
- **Statement 2**: True

### Final Answer
The correct option is that Statement 2 is correct and Statement 1 is incorrect.

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