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How to solve Q63

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Asked by tps.mjmdr 16th May 2019, 10:24 PM
Answered by Expert
Answer:
let masses of wire be m, 3m, 5m and their lengths be 5l, 3l, l.
 
if ρ is density of copper and a1, a2, a3 are cross section area of wires,
 
we have    m = ρ×a1×5l  .................(1)
                  3m = ρ×a2 ×3l ..............(2)
                  5m = ρ×a3×l  ................(3)
then ratio of a1 : a2 :a3 = (1/5) :1 : 5   or a1 : a2 :a3 = 1 : 5 : 25
 
hence we consider crosssection areas are a, 5a, 25a
 
Resistance ratio  = (5l/a) : (3l/5a) : (l/25a)   = 125 : 15 : 1
Answered by Expert 17th May 2019, 5:54 AM
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