A galvanometer of resistance 100 Ω has 50 divisions on its scale and has sensitivity of 20 μA/division. It is to be converted to voltmeter with three ranges, of 0 - 2 V, 0 - 10 V and 0 - 20 V. The appropriate circuit to do so is:

We have a moving-coil galvanometer whose resistance is $$R_g = 100\;\Omega$$. The scale has 50 divisions and the current sensitivity is $$20\;\mu\text{A}$$ per division.

First we find the full-scale (maximum) current of the galvanometer. Since each division needs $$20\;\mu\text{A}$$ and there are 50 divisions, we write

$$I_g = 50 \times 20\;\mu\text{A} = 50 \times 20 \times 10^{-6}\ \text{A} = 1000\;\mu\text{A} = 1\;\text{mA}.$$

To convert the galvanometer into a voltmeter, we keep the current through it limited to this same $$I_g = 1\;\text{mA}$$ while different external voltages are applied. For each desired range we first write Ohm’s law, $$V = I R,$$ then solve for the total resistance that must be in series with the galvanometer.

For the 0 - 2 V range, the total resistance required is

$$R_{2\text{V (total)}} = \dfrac{V}{I_g} = \dfrac{2\;\text{V}}{0.001\;\text{A}} = 2000\;\Omega.$$

This total already includes the galvanometer’s 100 Ω, so the series resistance needed is

$$R_{s1} = R_{2\text{V (total)}} - R_g = 2000 - 100 = 1900\;\Omega.$$

For the 0 - 10 V range, the total resistance required is

$$R_{10\text{V (total)}} = \dfrac{10\;\text{V}}{0.001\;\text{A}} = 10000\;\Omega,$$

and therefore the series resistance for this range alone must satisfy

$$R_{s(10)} = R_{10\text{V (total)}} - R_g = 10000 - 100 = 9900\;\Omega.$$

For the 0 - 20 V range, the total resistance required is

$$R_{20\text{V (total)}} = \dfrac{20\;\text{V}}{0.001\;\text{A}} = 20000\;\Omega,$$

so the required series resistance is

$$R_{s(20)} = R_{20\text{V (total)}} - R_g = 20000 - 100 = 19900\;\Omega.$$

In a multirange voltmeter the usual practice is to place resistors in series such that:

• Only the first resistor is used for the 2 V range,
• The first and the second together are used for the 10 V range,
• All three resistors are used for the 20 V range.

Let these resistors be $$R_1,\;R_2,\;R_3$$ connected sequentially with $$R_1$$ closest to the galvanometer. Then we require

$$\begin{aligned} R_1 &= 1900\;\Omega,\\ R_1 + R_2 &= 9900\;\Omega \;\;\Longrightarrow\;\; R_2 = 9900 - 1900 = 8000\;\Omega,\\ R_1 + R_2 + R_3 &= 19900\;\Omega \;\;\Longrightarrow\;\; R_3 = 19900 - 9900 = 10000\;\Omega. \end{aligned}$$

This gives the set $$R_1 = 1900\;\Omega,\; R_2 = 8000\;\Omega,\; R_3 = 10000\;\Omega,$$ with the ranges obtained in the order 2 V, 10 V, 20 V as required.

Comparing these values with the options, we see that they match exactly with Option D.

Hence, the correct answer is Option D.