JEE Mains · Physics · STD 12 - 3. current electricity
The length of a potentiometer wire is \(1200\; \mathrm{cm}\) and it carries a current of \(60 \;\mathrm{mA}\). For a cell of \(emf\;5\; \mathrm{V}\) and intemal resistance of \(20\; \Omega,\) the null point on it is found to be a \(1000\; \mathrm{cm} .\) The resistance of whole wire is .............. \(\Omega\)
- A \(120\)
- B \(60\)
- C \(80\)
- D \(100\)
Answer & Solution
Correct Answer
(D) \(100\)
Step-by-step Solution
Detailed explanation
\(5=\lambda \ell\) where \(\lambda\) is potential gradient and \(L\) is total length of wire. \(5=\frac{\Delta V}{L} \ell\) \(\Delta \mathrm{V}=\frac{5 \times \mathrm{L}}{\ell}=5 \times \frac{12}{10}=6 \mathrm{V}=60 \;\mathrm{mA} \times \mathrm{R}\) \(\mathrm{R}=100 \;\Omega\)
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