Mathematics in Physics

JEE Main level, single correct, medium

82 of 121

Question 82 of 121

A silver wire has a mass (0.6±0.006)g\left(0.6\pm 0.006\right)g, radius (0.5±0.005)mm\left(0.5\pm 0.005\right)mm and length (4±0.04)cm\left(4\pm 0.04\right)cm. The maximum percentage error in the measurement of its density will be

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Answer: (C) 44%

Volume of the wire is V=πr2lV=\pi {r}^{2}l.
Density of the wire is ρ=mπr2l\rho =\dfrac{m}{\pi {r}^{2}l}
Relative error in measurement of density
⇒Δρρ=Δmm+2×Δrr+Δll\Rightarrow \dfrac{\Delta \rho }{\rho }=\dfrac{\Delta m}{m}+2\times \dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}
=0.0060.6+2×0.0050.5+0.044=\dfrac{0.006}{0.6}+2\times \dfrac{0.005}{0.5}+\dfrac{0.04}{4}
=0.01+0.02+0.01=0.04=0.01+0.02+0.01=0.04
∴∴ Percentage error in ρ=0.04×100=4\rho =0.04\times 100=4%

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