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(D) Let $E$ be the electromotive force $(EMF)$ of the cell and $r_i$ be its internal resistance.The heat produced in a resistance $R$ is given by $H =

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$\sqrt{Rr}$

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(D) Let $E$ be the electromotive force $(EMF)$ of the cell and $r_i$ be its internal resistance.The heat produced in a resistance $R$ is given by $H = I^2 R t = \left(\frac{E}{R+r_i}\right)^2 R t$.According to the problem,the heat produced in both cases is the same for the same time $t$.Therefore,$\frac{E^2 R}{(R+r_i)^2} t = \frac{E^2 r}{(r+r_i)^2} t$.Canceling $E^2 t$ from both sides,we get $\frac{R}{(R+r_i)^2} = \frac{r}{(r+r_i)^2}$.Taking the square root on both sides: $\frac{\sqrt{R}}{R+r_i} = \frac{\sqrt{r}}{r+r_i}$.Cross-multiplying: $\sqrt{R}(r+r_i) = \sqrt{r}(R+r_i)$.$\sqrt{R}r + \sqrt{R}r_i = \sqrt{r}R + \sqrt{r}r_i$.Rearranging terms to solve for $r_i$: $r_i(\sqrt{R} - \sqrt{r}) = R\sqrt{r} - r\sqrt{R}$.$r_i(\sqrt{R} - \sqrt{r}) = \sqrt{Rr}(\sqrt{R} - \sqrt{r})$.Thus,$r_i = \sqrt{Rr}$.

$A$ cell sends a current through a resistance $R$ for time $t$. Now the same cell sends current through another resistance $r$ for the same time. If the same amount of heat is developed in both the resistances,then the internal resistance of the cell is:

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