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Current Electricity

JEE Main level, single correct, medium

360 of 380

Question 360 of 380

The galvanometer deflection, when key ${K}_{1}$ is closed but ${K}_{2}$ is open, equals ${\theta }_{0}$ (see figure). On closing ${K}_{2}$ also and adjusting ${R}_{2}$ to $5\Omega ,$ the deflection in galvanometer becomes $\dfrac{{\theta }_{0}}{5}.$ The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

Show answer and solution

Answer: (B) $22 \Omega$

{\theta ​}_{0}\propto {i}_{G}\Rightarrow {\theta ​}_{0}C={i}_{G}

I-case

C{\theta }_{0}=\dfrac{V}{220+R}

......(i)

II-case

C\dfrac{{\theta }_{0}}{5}=\dfrac{V}{\left(220+\dfrac{5R}{5+R}\right)}\times \dfrac{5}{5+R}

....(ii)

From (i) and (ii)

R=22\Omega

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JEE Main 2025 Apr 2 Shift 1 question paper

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