An unknown resistance R₁ is connected in series with a resistance of 10 Ω. This combination is connected to one gap of a metre bridge ; while a resistance R₂ is connected in the other gap. The balance point is at 50 cm. Now when 10 Ω resistance is removed, the balance point shifts by 10 cm. The value of R₁ is
Let initial balance condition be:
R₁ + 10/R₂ = 50/50
After removing 10 Ω, new balance condition:
R₁/R₂ = 40/60
Solving for R₁, we get 40 Ω. Answer: (c) 40 Ω.
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When the 10 Ω resistance is connected in series with the unknown resistance
R, the balance point is at 50 cm, indicating that the ratio of resistances in the two gaps is 1:1. Hence,
R₁ + 10 = R₂. After removing the 10 Ω resistance, the balance point shifts by 10 cm, making the new balance point 40 cm.
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