The correct option of this question will be (a).
Solution —
Current through galyanometer,
$I_{k}=\frac{V}{R+G}=\frac{2}{3000+25}=k \times 30$
where $k$ is figure of merit of galvanometer. The current corresponding 1020 units deflection of galvanometer:
$I=\frac{I_{g}}{30} \times 20=\frac{2}{3} I_{g}=\frac{2}{3} \times \frac{2}{3025} \mathrm{~A}$
If $R^{\prime}$ is the resistance to be used in series of galvanometer, then
$I=\frac{V}{R^{\prime}+G} \text { or } \frac{2}{3} \times \frac{2}{3025}=\frac{2}{R^{\prime}+25}$
On solving, we get
$R^{\prime}=4512.5 \Omega \approx 4513 \Omega$