Mathematics in Physics

JEE Main level, numerical answer, medium

74 of 121

Question 74 of 121

The length of a metallic wire is increased by 2020% and its area of cross-section is reduced by 44%. The percentage change in resistance of the metallic wire is ________.

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Answer: 25

The given data is
l′=1.2lA′=0.96Al'=1.2l A'=0.96A
Using the formula R=ρlAR=\dfrac{\rho l}{A}, the new resistance is
R′=1.2ρl0.96A{R}^{'}=\dfrac{1.2\rho l}{0.96A}
⇒R′=1.25R\Rightarrow {R}^{'}=1.25R
⇒ΔR=1.25R−R=0.25R\Rightarrow \Delta R=1.25R-R=0.25R
Hence, the percentage increase is
=ΔRR×100=0.25RR×100=25=\dfrac{\Delta R}{R}\times 100 =\dfrac{0.25R}{R}\times 100=25%

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