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A cell of emf $E$ 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 at the same time $t,$ the amount of heat generated in $R_{2}$ is $4 \mathrm{H}$. Then internal resistance of the cell is
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Verified Answer
The correct answer is:
$\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-2 \sqrt{R_{1}}}$
We have, $H=I^{2} R$ According to given condition, $I_{1}^{2} R_{1}=H$
and
$$
I_{2}^{2} R_{2}=4 H
$$
$$
\Rightarrow \quad \frac{E^{2}}{\left(R_{1}+r\right)^{2}} R_{1}=H \text { and } \frac{E^{2}}{\left(R_{2}+r\right)^{2}} R_{2}=4 H
$$
$\therefore \quad \frac{R_{2}}{\left(R_{2}+r\right)^{2}}=\frac{4 R_{1}}{\left(R_{1}+r\right)^{2}}$
$\Rightarrow \quad \sqrt{R_{2}}\left(R_{1}+r\right)=2 \sqrt{R_{1}}\left(R_{2}+r\right)$
On solving. $r=\frac{\sqrt{R_{1} R_{2}}\left[\sqrt{R_{1}}-2 \sqrt{R_{2}}\right]}{2 \sqrt{R_{1}}-\sqrt{R_{2}}}$
and
$$
I_{2}^{2} R_{2}=4 H
$$
$$
\Rightarrow \quad \frac{E^{2}}{\left(R_{1}+r\right)^{2}} R_{1}=H \text { and } \frac{E^{2}}{\left(R_{2}+r\right)^{2}} R_{2}=4 H
$$
$\therefore \quad \frac{R_{2}}{\left(R_{2}+r\right)^{2}}=\frac{4 R_{1}}{\left(R_{1}+r\right)^{2}}$
$\Rightarrow \quad \sqrt{R_{2}}\left(R_{1}+r\right)=2 \sqrt{R_{1}}\left(R_{2}+r\right)$
On solving. $r=\frac{\sqrt{R_{1} R_{2}}\left[\sqrt{R_{1}}-2 \sqrt{R_{2}}\right]}{2 \sqrt{R_{1}}-\sqrt{R_{2}}}$
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