Current Electricity

JEE Main level, numerical answer, medium

126 of 380

Question 126 of 380

Eight copper wire of length ll and diameter d are joined in parallel to form a single composite conductor of resistance RR. If a single copper wire of length 2l2l have the same resistance (R)\left(R\right) then its diameter will be _____ dd.

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Answer: 4

Each wire has resistance, r=ρ4lπd2r=\rho \dfrac{4l}{\pi {d}^{2}}
Eight wire in parallel, then equivalent resistance is
R=r8=ρl2πd2R=\dfrac{r}{8}=\dfrac{\rho l}{2\pi {d}^{2}}
Single copper wire of length 2l2l has resistance R=ρ2l×4πd12R=\rho \dfrac{2l\times 4}{\pi {d}_{1}^{2}}.
Comparing both equations, we get,
ρ8lπd12=ρl2πd2\rho \dfrac{8l}{\pi {d}_{1}^{2}}=\dfrac{\rho l}{2\pi {d}^{2}}
⇒d1=4d\Rightarrow {d}_{1}=4d

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