Science (086) — CBSE Class 10 board question

(a) The resistance of a uniform cylindrical conductor depends on:

1. Length (l): Resistance is directly proportional to length ($R \propto l$).
2. Area of cross-section (A): Resistance is inversely proportional to area ($R \propto 1/A$).
3. Nature of material (resistivity ρ): Different materials have different resistivities.

The relation is: $R = \rho \dfrac{l}{A}$

(b) Given: radius $r$ = 0.01 cm = $0.01 \times 10^{-2}$ m = $10^{-4}$ m, $R$ = 10 Ω, $\rho$ = 50 × 10⁻⁸ Ω m

Area $A = \pi r^2 = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8}$ m²

From $R = \rho\dfrac{l}{A}$:

$$l = \frac{RA}{\rho} = \frac{10 \times 3.14 \times 10^{-8}}{50 \times 10^{-8}} = \frac{31.4}{50} \approx 0.628 \text{ m}$$

The length of the wire is approximately 0.628 m.

Source: Chapter 11, Section 11.5

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• (a) is worth 2 marks — list all three factors clearly. The formula $R = \rho l/A$ ties them together and examiners expect it.
• (b) is worth 1 mark — show the substitution step clearly. Convert radius to metres first (common error: forgetting cm → m). The key formula is $l = RA/\rho$. Write the final answer with units.