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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
Options
(a) (1:3:5)
(b) (5:3:1)
(c) (1:15:25)
(d) (125:15:1)
Correct Answer:
(125:15:1)
Explanation:
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
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Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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