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$...\Omega$
A$78$
B$200$
C$2$
D$20$
JEE MAIN 2022, Diffcult
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D$20$
d Let resistance of potentiometers wire is $R$
$i=\frac{4}{R+780}$
Potential difference across $AB$
$=\frac{4 R}{R+780}$
Potential difference across $AC$
$=\frac{4 R \times 60}{(R+780) \times 300}=\frac{4 R}{5(R+780)}$
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