The resistance of a wire is given by the expression

$R = \frac{4 ρ L}{π D^{2}},$

where, $ρ$ is the resistivity $(i n O m e g a - m e t e r),$ L is the length (in meter) and D is the diameter (in meter) of the wire. The error in measurement of each of the parameters $ρ$ , L and D is $\pm 1.0 % .$ Assuming that the errors are independent random variables, the percent error in measurement of R is
2.45

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Solution

The correct option is A 2.45
$R = 4 \frac{ρ L}{π D^{2}}$

Assuming errors are in standard deviations.

$% σ_{ρ}, % σ_{L} a n d % σ_{D} = \pm 1 %$

$Δ R = \sqrt{{(\frac{\partial R}{\partial ρ} . Δ ρ)}^{2} + {(\frac{\partial R}{\partial L} . Δ L)}^{2} + {(\frac{\partial R}{\partial D} . Δ D)}^{2}}$

$% σ_{R} = \sqrt{{(% σ_{ρ})}_{ρ}^{2} + {(% σ_{L})}_{L}^{2} + {(2 \times % σ_{D})}_{D}^{2}}$

$\Rightarrow % σ_{R} = \sqrt{{(1)}^{2} + {(1)}^{2} + {(2 \times 1)}^{2}} = \sqrt{6}$

$\Rightarrow % σ_{R} = \pm \sqrt{6} = \pm 2.45 %$

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The correct option is A 2.45 R=4ρLπD2 Assuming errors are in standard deviations. % σρ,% σL and % σD=±1% ΔR=√(∂R∂ρ.Δρ)2+(∂R∂L.ΔL)2+(∂R∂D.ΔD)2 %σR=√(%σρ)2+(%σL)2+(2×%σD)2 ⇒ %σR=√(1)2+(1)2+(2×1)2=√6 ⇒ %σR=±√6=±2.45%