MHT CET · Physics · Current Electricity
A \(10 \mathrm{~m}\) long wire of resistance \(20 \Omega\) is connected in series with a battery of e.m.f.
3 volt and a resistance of \(10 \Omega\). The potential gradient along the wire in volt/meter
is
- A \(0 \cdot 02\)
- B \(1 \cdot 2\)
- C \(0 \cdot 10\)
- D \(0 \cdot 20\)
Answer & Solution
Correct Answer
(D) \(0 \cdot 20\)
Step-by-step Solution
Detailed explanation
(B)
Total resistance \(\mathrm{R}=20+10=30 \Omega\)
current I \(=\frac{\mathrm{E}}{\mathrm{R}}=\frac{3}{30}=0.1 \mathrm{~A}\)
p.d. acros wire \(V=I R_{w}\)
\(=0.1 \times 20=2 \mathrm{~V}\)
\(\begin{aligned} \therefore \text { Potential gradient } &=\frac{\mathrm{V}}{\mathrm{L}} \\ &=\frac{2}{10}=0.2 \mathrm{~V} / \mathrm{m} \end{aligned}\)
Total resistance \(\mathrm{R}=20+10=30 \Omega\)
current I \(=\frac{\mathrm{E}}{\mathrm{R}}=\frac{3}{30}=0.1 \mathrm{~A}\)
p.d. acros wire \(V=I R_{w}\)
\(=0.1 \times 20=2 \mathrm{~V}\)
\(\begin{aligned} \therefore \text { Potential gradient } &=\frac{\mathrm{V}}{\mathrm{L}} \\ &=\frac{2}{10}=0.2 \mathrm{~V} / \mathrm{m} \end{aligned}\)
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