A wire is stretched so as to change its diameter by ${\rm{0}}{\rm{.25}}\,\,{\rm{\% }}$. The percentage change in resistance is

$\begin{array}{*{20}{l}} {{\rm{1}}{\rm{.0}}\,\,{\rm{\% }}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{4}}{\rm{.0}}\,\,{\rm{\% }}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{2}}{\rm{.0}}\,\,{\rm{\% }}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.5}}\,\,{\rm{\% }}} \end{array}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

1 Medium

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Application +4 | -1

A wire is stretched so as to change its diameter by

\mathrm{0}\mathrm{.25}\,\,\mathrm{\% }

. The percentage change in resistance is

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {\mathrm{1}\mathrm{.0}\,\,\mathrm{\% }} \end{array}

Option C (1.0 %) is correct because the percentage change in resistance is calculated based on the change in diameter, considering constant volume.