First slide

Accuracy; precision of instruments and errors in measurements

Question

When a current of $(2.5 \pm 0.5) A$ flows through a wire, it develops a potential difference of $(20 \pm 1) V .$ The resistance of wire is

Moderate

A

$(8 \pm 2) Ω$
B

$(8 \pm 1.6) Ω$
C

$(8 \pm 1.5) Ω$
D

$(8 \pm 3) Ω$

Solution

We know resistance $R = \frac{V}{l} = \frac{20}{2.5} = 8 Ω$
$\frac{Δ R}{R} = \frac{Δ V}{V} + \frac{Δ l}{l} = \frac{1}{20} + \frac{0.5}{2.5} = \frac{1}{4}$
$Δ R = \frac{1}{4} \times 8 = 2 Ω = R (8 \pm 2) Ω$

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