TS EAMCET · Physics · Current Electricity
The area of cross-section of a potentiometer wire is \(6 \times 10^{-7} \mathrm{~m}^2\). The potential difference per unit length of the potentiometer wire when it is connected to a cell of negligible internal resistance and a resistor in series is \(0.15 \mathrm{Vm}^{-1}\). If the current through potentiometer wire is 0.3 A, then the resistivity of the material of the potentiometer wire is
- A \(4 \times 10^{-6} \Omega \mathrm{~m}\)
- B \(3 \times 10^{-7} \Omega \mathrm{~m}\)
- C \(3 \times 10^{-6} \Omega \mathrm{~m}\)
- D \(4 \times 10^{-7} \Omega \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(3 \times 10^{-7} \Omega \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\( \rho = \frac{kA}{I} \) \( \rho = \frac{(0.15)(6 \times 10^{-7})}{0.3} \) \( \rho = 3 \times 10^{-7} \Omega \mathrm{~m} \)
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