Electric Currents & Ohm's Law - Units of Resistivity Example 8
Question : The resistance of the wire used for telephone is 35 Ω per kilometer when the weight of the wire is 5 kg per kilometer. If the specific resistance of the material is 1.95 × 10−8 Ω-m, what is the cross-sectional area of the wire ? What will be the resistance of a loop to a subscriber 8 km from the exchange if wire of the same material but weighing 20 kg per kilometer is used ?
Solution :Here R = 35 Ω; l = 1 km = 1000 m; ρ = 1.95 × 10−8 Ω-m
Now,
R = ρl/A
= 55.7 × 10–8 m2
If the second case, if the wire is of the material but weighs 20 kg/km, then its cross-section must
be greater than that in the first case.
Cross-section in the second case = (20/5)x 55.7 x 10-8
= 222.8 x 10-8 m2
Length of wire = 2 × 8 = 16 km = 16000 m
∴ R = ρl/A
=140.1 Ω
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