A $10 \mathrm{~m}$ long wire of resistance $20 \Omega$ is connected in series with a battery of e.m.f. 3…

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
  1. $0 \cdot 02$
  2. $1 \cdot 2$
  3. $0 \cdot 10$
  4. $0 \cdot 20$

Solution

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 $V=I R_{w}$ $=0.1 \times 20=2 \mathrm{~V}$ $\begin{aligned} \therefore \text { Potential gradient } &=\frac{\mathrm{V}}{\mathrm{L}} \\ &=\frac{2}{10}=0.2 \mathrm{~V} / \mathrm{m} \end{aligned}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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