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A piece of wire of resistance `20 Omega` is drawn so that its length is increased to twice is original length. Calculate resistance of the wire in the new situation. |
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Answer» Correct Answer - `80 Omega` If the original length and the cross-sectional area of the wire are `l and A`, respectively, then `R = (rho l)/(A)` or `20 Omega = (rho l)/(A)`. When the length becomes `2 l` and cross-sectional area is `A//2`. `R_("new") = (rho(2 l))/(A//2) = 4(rho l//A) = 4(20 Omega) = 80 Omega`. |
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