A cell of e.m.f E is connected to a resistance $R_{1}$ for time t and the amount of heat generated in it is H. If the resistance $R_{1}$ is replaced by another resistance $R_{2}$ and is connected to the cell for the same time t, the amount of heat generated in $R_{2}$ is 4H. Then the internal resistance of the cell is

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$\frac{2 R_{1} + R_{2}}{2}$

$\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}} - \sqrt{R_{1}}}{\sqrt{R_{2}} - 2 \sqrt{R_{1}}}$

$\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}} - 2 \sqrt{R_{1}}}{2 \sqrt{R_{2}} - \sqrt{R_{1}}}$

$\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}} - \sqrt{R_{1}}}{2 \sqrt{R_{2}} - \sqrt{R_{1}}}$

answer is B.

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Detailed Solution

Heat produced in 1st resister

$H = i_{1}^{2} R_{1} t H = {(\frac{E}{R_{1} + r})}^{2} R_{1} t$

similarly $4 H = {(\frac{E}{R_{2} + r})}^{2} R_{2} t$

dividing $\frac{1}{4} = \frac{{(R_{2} + r)}^{2} R_{1}}{{(R_{1} + r)}^{2} R_{2}}$

taking square roots on both sides
$\begin{array}{l} \frac{1}{2} = \frac{(R_{2} + r) \sqrt{R_{1}}}{(R_{1} + r) \sqrt{R_{2}}} \\ R_{1} \sqrt{R_{2}} + r \sqrt{R_{2}} = 2 R_{2} \sqrt{R_{1}} + 2 r \sqrt{R_{1}} \\ r (\sqrt{R_{2}} - 2 \sqrt{R_{1}}) = 2 R_{2} \sqrt{R_{1}} - R_{1} \sqrt{R_{2}} \\ r = \frac{\sqrt{R_{1} R_{2}} (2 \sqrt{R_{2}} - \sqrt{R_{1}})}{\sqrt{R_{2}} - 2 \sqrt{R_{1}}} \end{array}$