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JEE Mains · Physics · STD 12 - 3. current electricity

As shown in the figure, a potentiometer wire of resistance \(20\,\Omega\) and length \(300\,cm\) is connected with resistance box (R.B.) and a standard cell of emf \(4\,V\). For a resistance ' \(R\) ' of resistance box introduced into the circuit, the null point for a cell of \(20\,mV\) is found to be \(60\,cm\). The value of ' \(R\) ' is \(.....\Omega\)

  1. A \(780\)
  2. B \(78\)
  3. C \(870\)
  4. D \(654\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(780\)

Step-by-step Solution

Detailed explanation

\(E =\frac{ AC }{ AB }\left( V _{ A }- V _{ B }\right)\) \(\therefore 20 \times 10^{-3}=\frac{60}{300} \times \frac{4 \times 20}{ R +20}\) \(\therefore R =780\,\Lambda\)
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