JAMB - Physics (1985 - No. 38)
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
8 : 1
4 : 1
2 : 1
1 : 1
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
\(\frac{R_Q}{R_P}\) = \(\frac{L_QA_P}{L_PA_Q}\)
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
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