A potentiometer wire has length $4 \mathrm{~m}$ and resistance $5 \Omega$. It is connected in series with…

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
  1. 0.03 V/m
  2. 0.01V/m
  3. $0.02 \mathrm{~V} / \mathrm{m}$
  4. $0.04 \mathrm{~V} / \mathrm{m}$

Solution

Total resistance $=495+5=500 \Omega$ current $\quad I=\frac{4}{500}=8 \times 10^{-3} \mathrm{~A}$ P.D. across the wire $=\mathrm{IR}$ $=8 \times 10^{-3} \times 5=40 \times 10^{-3} \mathrm{~V}$ $\begin{aligned} \therefore \text { Potential gradient } &=\frac{40 \times 10^{-3}}{4} \\ &=10^{-2} \mathrm{~V} / \mathrm{m}=0.01 \mathrm{~V} / \mathrm{m} \end{aligned}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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