A moving coil galvanometer, having a resistance $G$ produces full scales deflection when a current $I_{\mathrm{g}}$ flows through it. It can be converted into an ammeter of range $0$ to $I_0\left(>I_{\mathrm{g}}\right)$ by connecting a shunt of resistance $R_{\mathrm{A}}$ to it. If it can also be converted into a voltmeter of range $0$ to $V=I_{0}G$ by connecting a resistance $R_{\mathrm{V}}$ to it in series, then which among the following is correct?

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when $\mathrm{R.B.}=2400 \mathrm{\Omega }.$Deflection becomes 20 divisions when resistance taken from resistance box is 4900 Ω. Then we can conclude :



Note: This question is awarded as the bonus. Now the question is corrected.

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]: