First slide

Accuracy; precision of instruments and errors in measurements

Question

The specific resistance $ρ$ of a circular wire of radius r, resistance R, and length l is given by $ρ = π r^{2} R / l .$ Given: $r = 0.24 \pm 0.02 cm, R = 30 \pm 1 Ω, and l = 4.80 \pm 0.01 cm$ The percentage error in $ρ$ is nearly

Moderate

A

7%
B

9%
C

13%
D

20%

Solution

$\begin{matrix} Required percentage = (2 \frac{Δ r}{r} + \frac{Δ R}{R} + \frac{Δ l}{l}) 100 \\ = 2 \times \frac{0.02}{0.24} \times 100 + \frac{1}{30} \times 100 + \frac{0.01}{4.80} \times 100 \\ = 16.7 + 3.3 + 0.2 = 20 % \end{matrix}$

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