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A wire of uniform cross-section and length `l` has a resistance of `16 Omega ` is cut into four equal parts. Each part is stretched uniform to length `l` and all the four stretched parts are connected in parallel calcuate the total resistance of the combination so formed. Assume that stretching of wire does not cause any change in the density of its material |
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Answer» Correct Answer - `16 Omega` Resistance of the each of the four parts of length `l//4` is `R_(1) = (16)/(4) = 4 Omega` When each part is stretched to length `l`, its volume remains constant so `V = A_(1) ,l_(1) = A_(2) l_(2)or (A_(2))/(A_(1)) = (l_(1))/(l_(2)) = (1//4)/(l) = (1)/(4)` Now `R_(2) = rho 1//A or R prop l//A` `:.(R_(2))/(R_(1)) = (l_(2))/(l_(1)) xx (A_(1))/(A_(2)) = (l)/(1//4) xx 4 = 16` or `R_(1) = 16 R_(1) = 10xx 4 = 64 Omega` These resistance part `= 64 Omega` when these four are connected in parallel the effective resistance `R_("eff")` of the combination is given by `(1)/(R_(eff)) = (1)/(64) + (1)/(64) + (1)/(64) + (1)/(64) = (4)/(64) = (1)/(16)` or `R_(eff) = 16 Omega` |
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