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Question:

The galvanometer deflection, when key K1 is closed but K2 is open, equals θ₀ (see figure). On closing K2 also and adjusting R₂ to 5Ω, the deflection in the galvanometer becomes θ₀/5. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of the battery]: 12Ω, 5Ω, 22Ω, 25Ω

12Ω

25Ω

5Ω

22Ω

Solution:

The correct option is D 22Ω
case I ig = E/(220 + Rg) = Cθ₀. (i)
case II ig = (E/(220 + 5))/(5 + Rg) × 5/(Rg + 5) = Cθ₀/5.. (ii)
⇒ 5E/(225 + 1100) = Cθ₀/5. (ii)
E/(220 + Rg) = Cθ₀.. (i)
⇒ 225Rg + 1100/1100 + 5R₀ = 5
⇒ 5500 + 25Rg = 225Rg = 1100
200Rg = 4400
Rg = 22Ω