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# A wire P has half the diameter and half the length of a wire Q of similar materi...

### Question

A wire P has half the diameter and half the length of a wire Q of similar material. The ratio of the resistance of P to that of Q is

A) 8 : 1

B) 4 : 1

C) 2 : 1

D) 1 : 1

E) 1 : 4

### Explanation:

Resistance $$\alpha$$ $$\frac{length}{area}$$

$$\frac{R_Q}{R_P}$$ = $$\frac{L_QA_P}{L_PA_Q}$$

LQ = 2LP

DQ = 2DP

AP = $$\pi$$$$\frac{(2D_P)^2}{2}$$

= $$\pi$$DP

AQ = $$\pi$$$$\frac{(D_P)^2}{2}$$

= $$\frac{1}{4}$$$$\pi$$DP

therefore, $$\frac{R_Q}{R_P}$$ = $$\frac{2L_P \times \pi D_P}{L_P \times \frac{1}{4} \pi D_P}$$

= 2 : 1

## Dicussion (1)

• Resistance $$\alpha$$ $$\frac{length}{area}$$

$$\frac{R_Q}{R_P}$$ = $$\frac{L_QA_P}{L_PA_Q}$$

LQ = 2LP

DQ = 2DP

AP = $$\pi$$$$\frac{(2D_P)^2}{2}$$

= $$\pi$$DP

AQ = $$\pi$$$$\frac{(D_P)^2}{2}$$

= $$\frac{1}{4}$$$$\pi$$DP

therefore, $$\frac{R_Q}{R_P}$$ = $$\frac{2L_P \times \pi D_P}{L_P \times \frac{1}{4} \pi D_P}$$

= 2 : 1