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Two conductors are made of the same material and have the same length. Conductor `A` is a solid wire of diameter `1 mm`. Conductor `B` is a hollow tube of outer diameter `2 mm` and inner diameter `1mm`. Find the ratio of resistance `R_(A)` to `R_(B)`.
A. 3
B. 2
C. 1
D. 0.5

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Correct Answer - A
Resistance of wire, `R_(A)=(rho l)/(A)(rho l)/(pi[(D)/(2)]^(2))=(rho l)/((pi D^(2))/(4))`
`= (rho l)/(pi((1)^(2))/(4))=(4rho l)/(pi)`
Similarly, for hollow conductor B
`R_(B)=(rho l)/(A)=(rho l)/(pi[(D_(1)^(2))/(4)-(D_(2)^(2))/(4)])=(rho l)/(pi[(2^(2))/(4)-(1^(2))/(4)])=(4)/(3)(rho l)/(pi)`
`therefore " " (R_(A))/(R_(B))=(4rho l)/(pi)xx(3pi)/(4rho l)=3`

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