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A resistance wire conneted in the left gap of a metre brige balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is :
In a meter bridge, the balanced condition gives:
R/10 Ω = 3/2
Solving, R= 15Ω.
The resistance per unit length = 15Ω/1.5m = 10Ω/m.
Length of 1Ω = 1/10 = 0.1M = 1.0 × 10⁻¹ m.
Answer: (a) 1.0 × 10⁻¹m.
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