In a potentiometer, the area of cross section of the wire is $4 \mathrm{~cm}^2$, the current flowing in the…

In a potentiometer, the area of cross section of the wire is $4 \mathrm{~cm}^2$, the current flowing in the circuit is 1 A and the potential gradient is $7.5 \mathrm{Vm}^{-1}$, then the resistivity of the potentiometer wire is
  1. $3 \times 10^{-3} \Omega \mathrm{~m}$
  2. $2 \times 10^{-6} \Omega \mathrm{~m}$
  3. $4 \times 10^{-2} \Omega \mathrm{~m}$
  4. $5 \times 10^{-4} \Omega \mathrm{~m}$

Solution

In potentiometer, $\mathrm{A}=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$ $\mathrm{I}=1 \mathrm{~A}, \mathrm{k}=7.5 \mathrm{Vm}^{-1}$ $\therefore$ Potential gradient, $k=\frac{V}{l}=\frac{I_\delta}{A}$ $\therefore$ Resistivity $\delta=\frac{\mathrm{kA}}{\mathrm{I}}=\frac{7.5 \times 4 \times 10^{-4}}{1}$ $=3 \times 10^{-3} \Omega \mathrm{~m}$.

Asked in: AP EAMCET 2024 (18 May Shift 1)

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