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if the radius of cross section of the conductor is increased by 0.1% keeping volume constant, then percentage change in the resistance of the conductor is????

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  • 100
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r'= 1.001r                                where r is the initial radius of cross section of the wire
R = pl/A                                      since volume is constsnt   V= #r^2l        l is the length of the conductor
                                                     #(1.001r)^2l' = #r^2l
                                                                l' = l/ (1.001)^2
R' = pl'/#r'^2
    =   pl/(1.001)^2               
          #(1.001r)^2
    =   R
        (1.001)^4
    =  R
       1.004
% decrease in R  = R -  R     * 100  =  0.3%
                                    1.004
                                   R
i am hoping this is ryt
  • -39
But the answer is -0.4%
 
  • -10
Answer is -0.4 not -0.3
  • -2
Hi, Here is the answer Resistance = ρ * L /A Area increased by a factor = 1.001² = 1.002001 Length gets reduced by the same factor as the volume is constant = 1.002001 Resistance = ρ * L /A = ρ * (L/1.001002) / A * 1.001002 = [ρ * L /A ] * 1/1.002001² = Original Resistance 'R' * 0.996009 ≈ 0.996 * R The original resistance gets reduced by 0.4 %
  • 13
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