1. Key idea: charge on outer surface of each conducting shell

   For concentric conducting spherical shells, charges redistribute so that the electric field inside the metal is zero.

   Let the charges finally present on the outer surfaces of shells of radii $R,2R,3R$ be $q'_1,q'_2,q'_3$.

2. Shell 1 (radius $R$)

   Since there is no charge inside the innermost shell, all its charge stays on its outer surface: $q'_1 = Q_1$

3. Shell 2 (radius $2R$)

   The inner shell has charge $Q_1$, so to keep electric field zero inside the metal of shell 2, its inner surface gets charge $-Q_1$ Since total charge on shell 2 is $Q_2$, charge on its outer surface is $q'_2 = Q_2 - (-Q_1)=Q_1+Q_2$ More directly, $q'_2 = Q_1+Q_2$

4. Shell 3 (radius $3R$)

   Total charge enclosed inside shell 3 due to inner two shells is $Q_1+Q_2$ Hence its inner surface gets $-(Q_1+Q_2)$ Since total charge on shell 3 is $Q_3$, outer surface charge is $q'_3 = Q_3 + Q_1 + Q_2$ i.e. $q'_3 = Q_1+Q_2+Q_3$

5. Use equality of outer surface charge densities

   Surface charge density on outer surface is $\sigma = \frac{q'}{4\pi r^2}$

   Given these are equal for all three shells: $\frac{Q_1}{4\pi R^2} = \frac{Q_1+Q_2}{4\pi (2R)^2} = \frac{Q_1+Q_2+Q_3}{4\pi (3R)^2}$

   Cancel $4\pi R^2$: $Q_1 = \frac{Q_1+Q_2}{4} = \frac{Q_1+Q_2+Q_3}{9}$

6. Solve step by step

   From $Q_1 = \frac{Q_1+Q_2}{4}$ we get $4Q_1 = Q_1+Q_2$ $Q_2 = 3Q_1$

   From $Q_1 = \frac{Q_1+Q_2+Q_3}{9}$ we get $9Q_1 = Q_1+Q_2+Q_3$ $Q_3 = 8Q_1 - Q_2$ Substituting $Q_2=3Q_1$: $Q_3 = 8Q_1-3Q_1=5Q_1$

7. Therefore

   $Q_1:Q_2:Q_3 = 1:3:5$

8. Option check

   • A: $1:2:3$ ❌
   • B: $1:3:5$ ✅
   • C: $1:4:9$ ❌
   • D: $1:8:18$ ❌

So the correct answer is B.