Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths

Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths are in the ratio 3:2:1. The electric resistances are in the ratio

(a) (1:4:9)
(b) (9:4:1)
(c) (1:2:3)
(d) (27:6:1)

(27:6:1)

Mass = volume × density = A.l.d

Ald = m

Resistance R = ρl / A = ρl / (m/ld) = dρ (l²/m)

Now, R₂ / R₁ = (ρd / ρd) )(l₂² / l₁²) (m₁/m₂)

= (2/3)² × (1/2) = 2 / 9

⇒ R₂ = (2 / 9) R₁

Again R₃ / R₁ = (ρd / ρd) )(l₃² / l₁²) (m₁/m₃)

= (1/3)² × (1/3) = 1 / 27

⇒ R₃ = (R₁ / 27)

.·. R₁ : R₂ : R₃ = R₁ : (2/9) R₁ : (R₁/27) = 27 : 6 : 1

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