JEE Mains · Physics · STD 12 - 3. current electricity
An aluminium wire is stretched to make its length, \(0.4 \,\%\) larger. Then percentage change in resistance is \(.....\,\%\)
- A \(0.4\)
- B \(0.2\)
- C \(0.8\)
- D \(0.6\)
Answer & Solution
Correct Answer
(C) \(0.8\)
Step-by-step Solution
Detailed explanation
\(R =\frac{\rho \ell}{ A }\) \(\frac{\Delta R }{ R }=\frac{\Delta \ell}{\ell}-\frac{\Delta A }{ A }\) \(\ell A = k\) \(\frac{\Delta \ell}{\ell}+\frac{\Delta A }{ A }=0\) \(\frac{\Delta R }{ R }=\frac{2 \Delta \ell}{\ell}\) \(\frac{\Delta R }{ R }=2 \times 0.4=0.8 \%\)
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