The radius (r), length (l) and resistance (R) of a metal wire was measured in the laboratory as r = (0.35 ± 0.05) cm R = (100 ± 10) ohm l = (15 ± 0.2) cm The percentage error in resistivity of the material of the wire is: (a) 37.3% (b) 25.6% (c) 35.6% (d) 39.9%

Answer: (d). R = (ρ l)/(π r²) ⇒ ρ = (Rπ r²)/(l), so (Δρ)/(ρ) = (Δ R)/(R) + 2(Δ r)/(r) + (Δ l)/(l) (Δ R)/(R) = 10/100 = 10%, 2(Δ r)/(r) = 2·(0.05)/(0.35) = 28.57%, (Δ l)/(l) = 0.2/15 = 1.33% (Δρ)/(ρ) = 10 + 28.57 + 1.33 ≈ 39.9% So the answer is (d) 39.9%.

The radius (r), length (l) and resistance (R) of a metal wire was… | JEE Main 2024 (1 Feb, Shift 1)

Q58JEE Main 2024 · 1 FEB · SHIFT 1Easy

Errors in Measurement

The radius $(r)$ , length $(l)$ and resistance $(R)$ of a metal wire was measured in the laboratory as $r = (0.35 \pm 0.05)\text{ cm}$

$R = (100 \pm 10)\text{ ohm}$

$l = (15 \pm 0.2)\text{ cm}$

The percentage error in resistivity of the material of the wire is:

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Answer: (d)

$R = \frac{\rho l}{\pi r^2} \Rightarrow \rho = \frac{R\pi r^2}{l}$ , so $\frac{\Delta\rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$
$\frac{\Delta R}{R} = \frac{10}{100} = 10\%$ , $2\frac{\Delta r}{r} = 2\cdot\frac{0.05}{0.35} = 28.57\%$ , $\frac{\Delta l}{l} = \frac{0.2}{15} = 1.33\%$
$\frac{\Delta\rho}{\rho} = 10 + 28.57 + 1.33 \approx 39.9\%$
So the answer is (d) $39.9\%$ .

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