A current of ${\rm{0}}{\rm{.01}}\,\,{\rm{mA}}$ passes through the potentiometer wire of a resistivity of ${10^9}\,\,\,\,\Omega - {\rm{cm}}$ and area of cross-section ${10^{ - 2}}\,\,{\rm{c}}{{\rm{m}}^2}$. The potential gradient is

${10^{8}}\,\,{\rm{V}}{{\rm{m}}^{ - 1}}$

${10^9}\,\,{\rm{V}}{{\rm{m}}^{ - 1}}$

${10^{11}}\,\,{\rm{V}}{{\rm{m}}^{ - 1}}$

${10^{10}}\,\,{\rm{V}}{{\rm{m}}^{ - 1}}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

A current of

\mathrm{0}\mathrm{.01}\,\,\mathrm{mA}

passes through the potentiometer wire of a resistivity of

{10^9}\,\,\,\,\Omega - \mathrm{cm}

and area of cross-section

{10^{ - 2}}\,\,\mathrm{c}{\mathrm{m}^2}

. The potential gradient is

View answer and explanation

Correct option: D

{10^{8}}\,\,\mathrm{V}{\mathrm{m}^{ - 1}}

Option D (

{10^{8}}\,\mathrm{V}{\mathrm{m}^{ - 1}}

) is correct because it is the potential gradient calculated using the formula

k = \frac{{I\rho }}{A}