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

Question 1 Report

8 : 1

4 : 1

2 : 1

1 : 1

1 : 4

Answer Details

Resistance $α$ $\frac{l e n g t h}{a r e a}$

$\frac{R_{Q}}{R_{P}}$ = $\frac{L_{Q} A_{P}}{L_{P} A_{Q}}$

LQ = 2LP

DQ = 2DP

AP = $π$ $\frac{(2 D_{P})^{2}}{2}$

= $π$ DP

AQ = $π$ $\frac{(D_{P})^{2}}{2}$

= $\frac{1}{4}$ $π$ DP

therefore, $\frac{R_{Q}}{R_{P}}$ = $\frac{2 L_{P} \times π D_{P}}{L_{P} \times \frac{1}{4} π D_{P}}$

= 2 : 1

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