Ratio of Charges on Concentric Spherical Shells with Equal Outer Charge Densities

To find the ratio of the charges $Q_1 : Q_2 : Q_3$, we analyze the distribution of charges on the inner and outer surfaces of each concentric metallic spherical shell using electrostatic induction and Gauss's law.

Let the radii of the three concentric shells be:

• Innermost shell: $r_1 = R$
• Middle shell: $r_2 = 2R$
• Outermost shell: $r_3 = 3R$

Let the equal surface charge density on the outer surface of each shell be $\sigma$.

1. For the innermost shell of radius $R$:
   Since there is no charge in the cavity inside the innermost shell, the charge on its inner surface is zero:
   $q_{1,\text{inner}} = 0$
   Therefore, the entire charge $Q_1$ resides on its outer surface:
   $q_{1,\text{outer}} = Q_1$
   The surface charge density on the outer surface is:
   $\sigma = \frac{q_{1,\text{outer}}}{4\pi R^2} = \frac{Q_1}{4\pi R^2} \implies Q_1 = 4\pi R^2 \sigma$

2. For the middle shell of radius $2R$:
   By Gauss's law, the electric field inside the conducting material of the middle shell must be zero. Thus, the inner surface of this shell must induce a charge equal and opposite to the total enclosed charge:
   $q_{2,\text{inner}} = -Q_1$
   Since the net charge given to the middle shell is $Q_2$, by conservation of charge:
   $q_{2,\text{inner}} + q_{2,\text{outer}} = Q_2 \implies q_{2,\text{outer}} = Q_1 + Q_2$
   The surface charge density on its outer surface is:
   $\sigma = \frac{q_{2,\text{outer}}}{4\pi (2R)^2} = \frac{Q_1 + Q_2}{16\pi R^2}$
   Equating this to $\sigma = \frac{Q_1}{4\pi R^2}$:
   $\frac{Q_1 + Q_2}{16\pi R^2} = \frac{Q_1}{4\pi R^2}$
   $Q_1 + Q_2 = 4Q_1 \implies Q_2 = 3Q_1$

3. For the outermost shell of radius $3R$:
   Similarly, the charge enclosed by the inner surface of the outermost shell is the net charge on the inner two shells, which is $Q_1 + Q_2$. Thus, the induced charge on its inner surface is:
   $q_{3,\text{inner}} = -(Q_1 + Q_2)$
   Since the net charge given to this shell is $Q_3$, the charge on its outer surface is:
   $q_{3,\text{outer}} = Q_3 - q_{3,\text{inner}} = Q_1 + Q_2 + Q_3$
   The surface charge density on its outer surface is:
   $\sigma = \frac{q_{3,\text{outer}}}{4\pi (3R)^2} = \frac{Q_1 + Q_2 + Q_3}{36\pi R^2}$
   Equating this to $\sigma = \frac{Q_1}{4\pi R^2}$:
   $\frac{Q_1 + Q_2 + Q_3}{36\pi R^2} = \frac{Q_1}{4\pi R^2}$
   $Q_1 + Q_2 + Q_3 = 9Q_1$
   Substituting $Q_1 + Q_2 = 4Q_1$:
   $4Q_1 + Q_3 = 9Q_1 \implies Q_3 = 5Q_1$

Combining the results, we find:
$Q_1 : Q_2 : Q_3 = Q_1 : 3Q_1 : 5Q_1 = 1 : 3 : 5$

Therefore, the correct option is (B).