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((2 sigma)/(rho_(0)))`
C. `sqrt((rho_(0))/(2 sigma))`
D. `(rho_(0))/(sigma)`

Explanation

To find the ratio \( \frac{R_2}{R_1} \) for the given problem, we need to calculate the total charge of both the solid sphere and the hollow sphere and set their sum to zero.

### Step 1: Calculate the charge of the solid sphere

The volume charge density of the solid sphere is given by:
\[ \rho = \frac{\rho_0}{r} \]
To find the total charge \( Q_1 \) of the solid sphere, we will integrate the charge density over the volume of the sphere.

The differential charge \( dq \) in a thin shell of radius \( r \) and thickness \( dr \) is:
\[ dq = \rho \cdot dV = \rho \cdot (4\pi r^2 dr) = \frac{\rho_0}{r} \cdot (4\pi r^2 dr) = 4\pi \rho_0 r \, dr \]

Now, we integrate \( dq \) from \( r = 0 \) to \( r = R_1 \):
\[ Q_1 = \int_0^{R_1} 4\pi \rho_0 r \, dr \]
Calculating the integral:
\[ Q_1 = 4\pi \rho_0 \int_0^{R_1} r \, dr = 4\pi \rho_0 \left[ \frac{r^2}{2} \right]_0^{R_1} = 4\pi \rho_0 \cdot \frac{R_1^2}{2} = 2\pi \rho_0 R_1^2 \]

### Step 2: Calculate the charge of the hollow sphere

The hollow sphere has a surface charge density \( \sigma \) and a radius \( R_2 \). The total charge \( Q_2 \) on the hollow sphere is given by:
\[ Q_2 = \sigma \cdot A = \sigma \cdot (4\pi R_2^2) = 4\pi R_2^2 \sigma \]

### Step 3: Set the total charge to zero

According to the problem, the total charge in the system is zero:
\[ Q_1 + Q_2 = 0 \]
Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[ 2\pi \rho_0 R_1^2 + 4\pi R_2^2 \sigma = 0 \]

### Step 4: Rearranging the equation

Rearranging gives:
\[ 4\pi R_2^2 \sigma = -2\pi \rho_0 R_1^2 \]
Dividing both sides by \( 2\pi \):
\[ 2 R_2^2 \sigma = -\rho_0 R_1^2 \]
Now, we can express \( R_2^2 \) in terms of \( R_1^2 \):
\[ R_2^2 = -\frac{\rho_0 R_1^2}{2\sigma} \]

### Step 5: Finding the ratio \( \frac{R_2}{R_1} \)

Taking the square root of both sides:
\[ R_2 = R_1 \sqrt{-\frac{\rho_0}{2\sigma}} \]
Thus, the ratio \( \frac{R_2}{R_1} \) is:
\[ \frac{R_2}{R_1} = \sqrt{-\frac{\rho_0}{2\sigma}} \]

### Final Answer

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