Two wires P and Q are made of the same material. Wire P has twice the length and twice the diameter of wire Q. What is the ratio of resistance of P to resistance of Q?
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
Given:
- Length of P = 2l, Length of Q = l
- Diameter of P = 2d, Diameter of Q = d → Radius of P = 2r, Radius of Q = r
Using the formula $R = \rho \dfrac{l}{A}$, where $A = \pi r^2$:
$$R_P = \rho \cdot \frac{2l}{\pi (2r)^2} = \rho \cdot \frac{2l}{4\pi r^2} = \frac{\rho l}{2\pi r^2}$$
$$R_Q = \rho \cdot \frac{l}{\pi r^2}$$
$$\frac{R_P}{R_Q} = \frac{\rho l / 2\pi r^2}{\rho l / \pi r^2} = \frac{1}{2}$$
∴ Ratio of resistance of P to Q = 1 : 2
Source: Chapter 11 (Electricity), Section 11.5 – Factors on which the resistance of a conductor depends
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Explanation
- The key formula is $R = \rho \dfrac{l}{A}$, where $A = \pi r^2 = \pi(d/2)^2$.
- Doubling the length doubles R, but doubling the diameter quadruples the area, which quarters R. Net effect: $2 \times \frac{1}{4} = \frac{1}{2}$.
- Examiners expect you to substitute values clearly and show the cancellation step. Writing the final ratio in the form 1 : 2 is mandatory.