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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

Options

A) 8 : 1

B) 4 : 1

C) 2 : 1

D) 1 : 1

E) 1 : 4


The correct answer is C.

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


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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

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