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MHT CET · Physics · Current Electricity

A potentiometer wire of length \(4 \mathrm{~m}\) and resistance \(5 \Omega\) is connected in series with a resistance of \(992 \Omega\) and a cell of e.m.f. \(4 \mathrm{~V}\) with internal resistance \(3 \Omega\). The length of \(0.75 \mathrm{~m}\) on potentiometer wire balances the e.m.f. of

  1. A \(4.00 \mathrm{mV}\)
  2. B \(3.75 \mathrm{mV}\)
  3. C \(3.00 \mathrm{mV}\)
  4. D \(2.50 \mathrm{mV}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3.75 \mathrm{mV}\)

Step-by-step Solution

Detailed explanation

\(\therefore \quad\) Total Resistance:
\(
\mathrm{R}=992+5+3=1000 \Omega
\)
Voltage across \(4 \mathrm{~m}\) wire:
\(
\frac{5}{995+5} \times 4=0.02 \mathrm{~V}
\)
\(\therefore \quad\) For one metre wire:
\(
\frac{0.02}{4}=0.005 \mathrm{~V}
\)
\(\therefore \quad\) For \(0.75 \mathrm{~m}\) wire:
\(
\begin{aligned}
0.004 \times 0.75 & =0.00375 \\
& =3.75 \mathrm{mV}
\end{aligned}
\)