The masses of three wires of copper are in the ratio 1:3:5 and lengths

The masses of three wires of copper are in the ratio 1:3:5 and lengths are in the ratio 5:3:1. Then the ratio of their electrical resistance are

(a) (1:3:5)
(b) (5:3:1)
(c) (1:15:25)
(d) (125:15:1)

(125:15:1)

Let A₁, A₂ and A₃ be the area of cross-section of three wires of copper of masses m₁,m₂ and m₃ and length l₁,l₂ and l₃ respectively.

m₂ = 3m, m₃ = 5m, l₁ = 3l, l₃ = l

Since Mass = Volume density

.·. m = A₁ × 5l × ρ —-(i)

3m = A₂ × 3l × ρ —(ii)

5m = A₃ × 1l × ρ —(iii)

From (i) and (iii), we get

A₂ = 5A₁

From (i) and (iii), we get

A₃ = 25 A₁

.·. R₁ = ρl₁ / A₁ = ρ5l / A₁, R₂ = ρl₂ / A₂ = ρ3l / A₁ = (3 / 25) R₁

R₃ = ρl₃ / A₃ = ρl / 25 A₁ = (R₁ / 125) = 125 : 15 : 1

admin: