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  • The resistivity of a potentiometer wire is <img alt="\mathrm{40\times 10^{-8} \Omega m }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B40%5Ctimes%2010%5E%7B-8%7D%20%5COmega%20m%

The resistivity of a potentiometer wire is \mathrm{40\times 10^{-8} \Omega m }  and its area of cross – section is \mathrm{ 8\times 10^{-6} m ^{2}}. If \mathrm{0.2 A} current is flowing through the wire, the potential gradient will be:

Option: 1

10^{-2}\mathrm{m}^{-1}


Option: 2

10^{-1}\mathrm{V\: m}^{-1}


Option: 3

3.2\times10^{-2}\mathrm{Vm}^{-1}


Option: 4

1\mathrm{Vm}^{-1}


Answers (1)

best_answer

The resistance of 1 \mathrm{~m} length wire will be \mathrm{R}=\frac{\left(40 \times 10^{-8}\right)}{\left(8 \times 10^{-6}\right)}=5 \times 10^{-2} \Omega

Potential gradient,

\begin{aligned} & \frac{\Delta \mathrm{V}}{\Delta \mathrm{x}}=\frac{\mathrm{R} . \mathrm{I}}{l} \\ \end{aligned}

\begin{aligned} & \frac{\Delta \mathrm{V}}{\Delta \mathrm{x}}=\left(5 \times 10^{-2}\right) \times 0.2=10^{-2} \mathrm{Vm}^{-1} \end{aligned}

Hence option 1 is correct.

Posted by

Rishabh

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