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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)`. |
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Answer» For a solid wire of resistance `R_(A^(,)) l_(1) = l, p_(1) = p, D_(1) = 1 mm, A_(1) = piD_(1)^(2)//4 = pi(1)^(2)//4mm^(2)` `R_(A) = (p_(1)l_(1))/A_(1) = (pl)/(pi(1)^(2)//4) = (4pl)/pi` For hollow tube of resistance `R_(B), l_(2) = l, p_(2) = p, A_(2) = pi(D_(2)^(2)-D_(1)^(2))//4 = pi(2^(2)-1^(2))//4 = 3 pi//4 mm^(2)` `R_(B) = (p_(2)l_(2))/A_(2) = (pl)/((3pi//4)) = (4pl)/(3pi) :. R_(A)//R_(B) = 3` |
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