The masses of three wires of copper are in the ratio 1:3:5, and their lengths are in the ratio of 5:3:1. The ratio of their electrical resistances are _______.

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Solution

This is a typical question on resistance formula:
Given:
$M_{1} : M_{2} : M_{3} = 1 : 3 : 5$
$l_{1} : l_{2} : l_{3} = 5 : 3 : 1$
We know that:
R = ρ(l/A)
Mass = Volume
Density.
Mass = A l d
⇒ A = M/(l d)
Replacing value of A in resistance formula

We get:
$R = ρ (l^{2}) d / M$
Density and resistivity is the same as all three wires of copper
$R_{1} : R_{2} : R_{3} =$
⇒ $(l_{1})^{2} / (M_{1}) : (l_{2})^{2} / (M_{2}) : (l_{3})^{2} / (M_{3})$
⇒(25/1):(9/3):(1/5)
⇒125:15:1

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The masses of three wires of copper are in the ratio 1:3:5, and their lengths are in the ratio of 5:3:1. The ratio of their electrical resistances are _______.​

The masses of three wires of copper are in the ratio 1:3:5, and their lengths are in the ratio of 5:3:1. The ratio of their electrical resistances are _______.