The galvanometer deflection, when key $\mathrm{K}_{1}$ is closed but $\mathrm{K}_{2}$ is open, equals $\theta_{0}$ (see figure). On closing $\mathrm{K}_{2}$ 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]:
Correct Option: , 2
(2) When key $\mathrm{K}_{1}$ is closed and key $\mathrm{K}_{2}$ is open
$\mathrm{i}_{\mathrm{g}}=\frac{\mathrm{E}}{220+\mathrm{R}_{\mathrm{g}}}=\mathrm{C} \theta_{0}$ ...(1)
When both the keys are closed
$\mathrm{i}_{\mathrm{g}}=\left(\frac{\mathrm{E}}{220+\frac{5 \mathrm{R}_{\mathrm{g}}}{5+\mathrm{R}_{\mathrm{g}}}}\right) \times \frac{5}{\left(\mathrm{R}_{\mathrm{g}}+5\right)}=\frac{\mathrm{C} \theta_{0}}{5}$
$\Rightarrow \frac{5 \mathrm{E}}{225 \mathrm{R}_{\mathrm{g}}+1100}=\frac{\mathrm{C} \theta_{0}}{5}$ ...(2)
$\frac{\mathrm{E}}{220+\mathrm{R}_{\mathrm{g}}}=\mathrm{C} \theta_{0}$ ...(1)
Dividing (i) by (ii), we get
$\Rightarrow \frac{225 \mathrm{R}_{\mathrm{g}}+1100}{1100+5 \mathrm{R}_{\mathrm{g}}}=5$
$\Rightarrow 5500+25 R_{\mathrm{g}}=225 R_{\mathrm{g}}+1100$
$200 R_{\mathrm{g}}=4400$
$R_{\mathrm{g}}=22 \Omega$
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