Generated by claude-sonnet-4-6 · 2026-06-15 06:47 · grounding rag
Model Answer
(i) Resistance of a Conductor & Its SI Unit
The resistance of a conductor is its property of opposing the flow of electric current through it. It is defined by Ohm's law as:
$$R = \frac{V}{I}$$
SI unit: The SI unit of resistance is ohm (Ω). One ohm is the resistance of a conductor when a potential difference of 1 volt drives a current of 1 ampere through it.
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(ii) Two Factors on Which Resistance Depends
- Length (l): Resistance is directly proportional to the length of the conductor ($R \propto l$).
- Area of cross-section (A): Resistance is inversely proportional to the area of cross-section ($R \propto \frac{1}{A}$).
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(iii) Effect on Resistance When Length and Radius Are Both Doubled
Using $R = \rho\dfrac{l}{A}$, where $A = \pi r^2$:
- Length doubled ($l \to 2l$): $R \propto l$, so resistance doubles.
- Radius doubled ($r \to 2r$): $A = \pi(2r)^2 = 4\pi r^2$, so area becomes 4 times. Since $R \propto \frac{1}{A}$, resistance becomes one-fourth.
Net effect: $R_{\text{new}} = \rho\dfrac{2l}{4A} = \dfrac{1}{2}R$
The resistance becomes half the original value.
Source: Chapter 11, Section 11.5 – Factors on which resistance depends
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Explanation
- (i) Always define resistance AND state the SI unit with the condition (1V, 1A → 1Ω). Examiners expect both.
- (ii) The textbook lists length and area of cross-section as the two main factors (nature of material is the third). Pick any two; length and area are safest.
- (iii) The key formula is $R = \rho\frac{l}{A}$. Remember that doubling the radius quadruples the area ($A = \pi r^2$). Combine both changes to get the final answer of R/2. Show the calculation step for full marks.