Question

Masses of three wires of copper are in the ratio 5 : 3 : 1 and their lengths are in the ratio 1 : 3 : 5. The ratio of their electrical resistances are _____ (A) 5 : 3 : 1
(B) √125 : 15 : 1
(C) 1 : 15 : 125
(D) 1 : 3 : 5

Answer 1

Given that mass ratio of 3 wires 1 : 3 : 5 , lengths are in ratio 5 : 3: 1,

V = A * L, V = pi*r^2 * L,

density = d,

m = V*d,

m = A*L * d , A = m/(L *d), R = p*L/A, p is constant for all copper so now we just have to find the ratio of L to area, since d = density of copper it is constant too ..

so the only variables are , A varies m/L, R varies L/A, R varies by L^2/m, first wire R1 = 5^2/1 = 25, R2 = 3^2/3 = 3, R3 = 1^2/5 = 1/5, the resistance vary by 25 : 3: 1/5.

Hence it will be more comfortable for the student to get an idea on solving this answers.