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\(...\Omega\)

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Answer & Solution

Correct Answer

(D) \(20\)

Detailed explanation

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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