Mathematics in Physics

JEE Main level, single correct, medium

104 of 121

Question 104 of 121

A copper wire is stretched to make it 0.50.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is:

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Answer: (B) 1.01.0%

Volume, V=A.lV=A.l
In differential form,
dVV=dAA+dll0=dAA+dll\dfrac{dV}{V}=\dfrac{dA}{A}+\dfrac{dl}{l} 0=\dfrac{dA}{A}+\dfrac{dl}{l}

⇒dAA=−dll\Rightarrow \dfrac{dA}{A}=-\dfrac{dl}{l}Also resistance of wire is as follows R=ρlAR=\rho \dfrac{l}{A}
diffrentiation formula for resistance is as follows
dRR=dll−dAAdRR=dll−(−dll)\dfrac{dR}{R}=\dfrac{dl}{l}-\dfrac{dA}{A} \dfrac{dR}{R}=\dfrac{dl}{l}-(-\dfrac{dl}{l})
dRR=2⋅dll\dfrac{dR}{R}=2\cdot \dfrac{dl}{l}
dRR×100=2⋅dll×100dRR\dfrac{dR}{R}\times 100=2\cdot \dfrac{dl}{l}\times 100 \dfrac{dR}{R}%=2\times 0.5% \dfrac{dR}{R}%=1%

∴∴% change in R=2×0.5R=2\times 0.5%=1%

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