MHT CET · Physics · Current Electricity
The resistivity of potentiometer wire is \(40 \times 10^{-8} \Omega m\) and its area of cross-section is 8 \(\times 10^{-6} m^2\). If 0.2 A current is flowing through the wire, the potential gradient of the wire is
- A \(0.1\text{ V m} ^{-1}\)
- B \(10^{-2}\text{ V m} ^{-1}\)
- C \(10^{-3}\text{ V m} ^{-1}\)
- D \(10^{-4}\text{ V m} ^{-1}\)
Answer & Solution
Correct Answer
(B) \(10^{-2}\text{ V m} ^{-1}\)
Step-by-step Solution
Detailed explanation
\(R=\frac{\rho l}{A}\)
\(\therefore \frac{R}{l}=\frac{\rho}{A}=\frac{40 \times 10^{-8}}{8 \times 10^{-6}}=5 \times 10^{-2}\)
\(\frac{V}{l}=\frac{I R}{l}=0.2 \times 5 \times 10^{-2}=10^{-2}\text{ V m} ^{-1}\)
\(\therefore \frac{R}{l}=\frac{\rho}{A}=\frac{40 \times 10^{-8}}{8 \times 10^{-6}}=5 \times 10^{-2}\)
\(\frac{V}{l}=\frac{I R}{l}=0.2 \times 5 \times 10^{-2}=10^{-2}\text{ V m} ^{-1}\)
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