A potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is$...\Omega$
JEE MAIN 2022, Diffcult
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Let resistance of potentiometers wire is $R$
$i=\frac{4}{R+780}$
Potential difference across $AB$
$=\frac{4 R}{R+780}$
Potential difference across $AC$
$=\frac{4 R \times 60}{(R+780) \times 300}=\frac{4 R}{5(R+780)}$
This should be equal to $20\,mV$
$\frac{4 R}{5(R+780)}=20 \times 10^{-3}=2 \times 10^{-2}$
$4 R=10^{-1}(R+780)$
$4 R=\frac{R}{10}+78$
$4 R-\frac{R}{10}=78$
$\frac{39 R}{10}=78$
$R=20\,\Omega$
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