JEE Mains · Physics · STD 12 - 3. current electricity
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 \(\frac{{\theta _0}}{5}\). The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: .................. \(\Omega\)

- A \(5\)
- B \(22\)
- C \(25\)
- D \(12\)
Answer & Solution
Correct Answer
(B) \(22\)
Step-by-step Solution
Detailed explanation
When \(\mathrm{K}_{1}\) closed and \(\mathrm{K}_{2}\) is open \(i=\frac{V}{220+R}\) .....\((i)\) When \(\mathrm{k}_{1}\) and \(\mathrm{k}_{2}\) are closed. \(\mathrm{KVL} \rightarrow\)…
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