MHT CET · Physics · Current Electricity
A potentiometer wire has length \(4 \mathrm{~m}\) and resistance \(5 \Omega\). It is connected in series with
\(495 \Omega\) resistance and a cell of e.m.f. \(4 \mathrm{~V}\). The potential gradient along the wire is
- A 0.03 V/m
- B 0.01V/m
- C \(0.02 \mathrm{~V} / \mathrm{m}\)
- D \(0.04 \mathrm{~V} / \mathrm{m}\)
Answer & Solution
Correct Answer
(B) 0.01V/m
Step-by-step Solution
Detailed explanation
Total resistance \(=495+5=500 \Omega\)
current \(\quad I=\frac{4}{500}=8 \times 10^{-3} \mathrm{~A}\)
P.D. across the wire \(=\mathrm{IR}\)
\(=8 \times 10^{-3} \times 5=40 \times 10^{-3} \mathrm{~V}\)
\(\begin{aligned} \therefore \text { Potential gradient } &=\frac{40 \times 10^{-3}}{4} \\ &=10^{-2} \mathrm{~V} / \mathrm{m}=0.01 \mathrm{~V} / \mathrm{m} \end{aligned}\)
current \(\quad I=\frac{4}{500}=8 \times 10^{-3} \mathrm{~A}\)
P.D. across the wire \(=\mathrm{IR}\)
\(=8 \times 10^{-3} \times 5=40 \times 10^{-3} \mathrm{~V}\)
\(\begin{aligned} \therefore \text { Potential gradient } &=\frac{40 \times 10^{-3}}{4} \\ &=10^{-2} \mathrm{~V} / \mathrm{m}=0.01 \mathrm{~V} / \mathrm{m} \end{aligned}\)
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