A cell of constant emf is first connected to a resistance $R_1$ and then connected to a resistance $R_2$. If power delivered in both cases is same, then the internal resistance of the cell is

$\sqrt{R_1R_2}$

The correct answer is Option (1) → $\sqrt{R_1R_2}$

Let emf of the cell = $E$, internal resistance = $r$

Power across $R = \frac{E^2 R}{(R+r)^2}$

Given: Power across $R_1$ = Power across $R_2$

$\frac{E^2 R_1}{(R_1 + r)^2} = \frac{E^2 R_2}{(R_2 + r)^2}$

Cancel $E^2$:

$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$

Cross-multiply:

$R_1(R_2 + r)^2 = R_2(R_1 + r)^2$

Expanding:

$R_1(R_2^2 + 2R_2r + r^2) = R_2(R_1^2 + 2R_1r + r^2)$

$R_1R_2^2 + 2R_1R_2r + R_1r^2 = R_2R_1^2 + 2R_1R_2r + R_2r^2$

Cancel $2R_1R_2r$:

$R_1R_2^2 + R_1r^2 = R_2R_1^2 + R_2r^2$

Rearrange:

$R_1R_2^2 - R_2R_1^2 = R_2r^2 - R_1r^2$

$R_1R_2(R_2 - R_1) = r^2(R_2 - R_1)$

If $R_1 \neq R_2$, divide by $(R_2 - R_1)$:

$R_1R_2 = r^2$

$r = \sqrt{R_1R_2}$

Final Answer: $r = \sqrt{R_1R_2}$