A cell of e.m.f E is connected to a resistance R1 for time t and the amount of heat generated in it is H. If the resistance R1 is replaced by another resistance R2 and is connected to the cell for the same time t, the amount of heat generated in R2 is 4H. Then the internal resistance of the cell is

R 1 R 2 R 2 − 2 R 1 2 R 2 − R 1

A cell of e.m.f E is connected to a resistance R1 for time t and the amount of heat generated in it is H. If the resistance R1 is replaced by another resistance R2 and is connected to the cell for the same time t, the amount of heat generated in R2 is 4H. Then the internal resistance of the cell is

R 1 R 2 R 2 − 2 R 1 2 R 2 − R 1

R 1 R 2 2 R 2 − R 1 R 2 − 2 R 1

R 1 R 2 R 2 − 2 R 1 2 R 2 − R 1

R 1 R 2 R 2 − R 1 R 2 − R 1

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A cell of e.m.f E is connected to a resistance R₁for time t and the amount of heat generated in it is H. If the resistance R₁ is replaced by another resistance R₂and is connected to the cell for the same time t, the amount of heat generated in R₂ is 4H. Then the internal resistance of the cell is

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

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

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

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

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
$\begin{array}{l} \frac{1}{2} = \frac{(R_{2} + r) \sqrt{R_{1}}}{(R_{1} + r) \sqrt{R_{2}}} \\ R_{1} \sqrt{R_{2}} + r \sqrt{R_{2}} = 2 R_{2} \sqrt{R_{1}} + 2 r \sqrt{R_{1}} \\ r (\sqrt{R_{2}} - 2 \sqrt{R_{1}}) = 2 R_{2} \sqrt{R_{1}} - R_{1} \sqrt{R_{2}} \\ r = \frac{\sqrt{R_{1} R_{2}} (2 \sqrt{R_{2}} - \sqrt{R_{1}})}{\sqrt{R_{2}} - 2 \sqrt{R_{1}}} \end{array}$

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