A solid sphere of radius `R_(1)` and volume charge density `rho = (rho_(0))/(r )` is enclosed by a hollow sphere of radius `R_(2)` with negative surface charge density `sigma`, such that the total charge in the system is zero . `rho_(0)` is positive constant and `r` is the distance from the centre of the sphere . The ratio `R_(2)//R_(1)` is

Options

A. `sigma/rho_0`
B. `sqrt((2sigma)/(rho_0))`
C. `sqrt((rho_0)/(2sigma))`
D. `rho_0/sigma`

Answer

To solve the problem, we need to find the ratio \( \frac{R_2}{R_1} \) given a solid sphere of radius \( R_1 \) with a volume charge density \( \rho = \frac{\rho_0}{r} \) and a hollow sphere of radius \( R_2 \) with a negative surface charge density \( \sigma \) such that the total charge in the system is zero.

### Step-by-Step Solution:

1. **Identify the Charge of the Solid Sphere \( Q_1 \)**:
- The charge \( Q_1 \) in the solid sphere can be calculated by integrating the volume charge density over the volume of the sphere.
- The volume charge density is given as \( \rho = \frac{\rho_0}{r} \).
- The charge element \( dQ \) for a thin spherical shell of radius \( r \) and thickness \( dr \) is:
\[ dQ = \rho \cdot dV = \rho \cdot (4\pi r^2 dr) = \frac{\rho_0}{r} \cdot (4\pi r^2 dr) = 4\pi \rho_0 r dr \]
- To find the total charge \( Q_1 \), integrate from \( 0 \) to \( R_1 \):
\[ Q_1 = \int_0^{R_1} 4\pi \rho_0 r \, dr = 4\pi \rho_0 \int_0^{R_1} r \, dr = 4\pi \rho_0 \left[ \frac{r^2}{2} \right]_0^{R_1} = 4\pi \rho_0 \frac{R_1^2}{2} = 2\pi \rho_0 R_1^2 \]

2. **Identify the Charge of the Hollow Sphere \( Q_2 \)**:
- The charge \( Q_2 \) on the hollow sphere is given by the surface charge density \( \sigma \):
\[ Q_2 = -\sigma \cdot A = -\sigma \cdot (4\pi R_2^2) = -4\pi \sigma R_2^2 \]

3. **Set Up the Equation for Total Charge**:
- Since the total charge in the system is zero:
\[ Q_1 + Q_2 = 0 \]
- Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[ 2\pi \rho_0 R_1^2 - 4\pi \sigma R_2^2 = 0 \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ 2\pi \rho_0 R_1^2 = 4\pi \sigma R_2^2 \]
- Dividing both sides by \( 2\pi \):
\[ \rho_0 R_1^2 = 2\sigma R_2^2 \]

5. **Finding the Ratio \( \frac{R_2}{R_1} \)**:
- Rearranging the equation for \( R_2 \):
\[ \frac{R_2^2}{R_1^2} = \frac{\rho_0}{2\sigma} \]
- Taking the square root gives:
\[ \frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}} \]

### Final Answer:
The ratio \( \frac{R_2}{R_1} \) is:
\[ \frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}} \]