Mathematics in Physics

JEE Main level, single correct, medium

68 of 121

Question 68 of 121

The radius (r)\left(r\right), length (l)\left(l\right) and resistance (R)\left(R\right) of a metal wire was measured in the laboratory as
r=(0.35±0.05)cmr=\left(0.35\pm 0.05\right)cm, R=(100±10)ohmR=\left(100\pm 10\right)ohm, l=(15±0.2)cml=\left(15\pm 0.2\right)cm
The percentage error in resistivity of the material of the wire is :

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Answer: (B) 39.939.9%

The resistance of the wire is given as, R=ρlAR=\dfrac{\rho l}{A}
Therefore, ρ=RAl\rho =R\dfrac{A}{l}
ρ=R(πr2)l\rho =R\dfrac{\left(\pi {r}^{2}\right)}{l}
For percentage error, we can write Δρρ=ΔRR+2Δrr+Δll\dfrac{\Delta \rho }{\rho }=\dfrac{\Delta R}{R}+2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}
Δρρ×100=ΔRR×100+2Δrr×100+Δll×100\dfrac{\Delta \rho }{\rho }\times 100=\dfrac{\Delta R}{R}\times 100+2\dfrac{\Delta r}{r}\times 100+\dfrac{\Delta l}{l}\times 100
=(10100+2×0.050.35+0.215)×100=\left(\dfrac{10}{100}+2\times \dfrac{0.05}{0.35}+\dfrac{0.2}{15}\right)\times 100
=(110+27+175)×100=\left(\dfrac{1}{10}+\dfrac{2}{7}+\dfrac{1}{75}\right)\times 100
Hence, Δρρ=39.9\dfrac{\Delta \rho }{\rho }=39.9%

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