A galvanometer of resistance $$G$$ is converted into a voltmeter of range $$0 - 1\,\text{V}$$ by connecting a resistance $$R$$ in series with it. The additional resistance that should be connected in series with $$R_1$$ to increase the range of the voltmeter to $$0 - 2\,\text{V}$$ will be:

To begin, let us denote the resistance of the galvanometer by $$G$$ and its full-scale deflection current by $$I_g$$. When we wish to read a voltage higher than the galvanometer can withstand directly, we place a resistance in series so that the same current $$I_g$$ will flow at that higher voltage.

First range (0 - 1 V). We are told that a series resistance $$R_1$$ (the statement first calls it $$R$$, then labels it $$R_1$$; we shall use $$R_1$$) converts the galvanometer into a voltmeter whose maximum reading is $$1\;\text{V}$$. By Ohm’s law we must have

$$I_g \bigl(G + R_1\bigr) \;=\; 1$$

because at full-scale deflection the total resistance seen by the current is the sum $$G + R_1$$ and the corresponding voltage is $$1\;\text{V}$$.

We can solve this for $$I_g$$, since we shall need it later:

$$I_g = \dfrac{1}{G + R_1} \qquad\bigl[\text{Equation (1)}\bigr]$$

Second range (0 - 2 V). We now want the same galvanometer to read up to $$2\;\text{V}$$. Let the total series resistance required for this new range be $$R_{\text{total}}^{(2)}$$. Again applying Ohm’s law at full-scale, we set

$$I_g \bigl(G + R_{\text{total}}^{(2)}\bigr) \;=\; 2$$

Solving for $$R_{\text{total}}^{(2)}$$ gives

$$R_{\text{total}}^{(2)} = \frac{2}{I_g} - G \qquad\bigl[\text{Equation (2)}\bigr]$$

Substituting for $$I_g$$. From Equation (1) we know $$I_g = 1/(G + R_1)$$, so

$$\frac{2}{I_g} = 2\bigl(G + R_1\bigr)$$

Putting this into Equation (2):

$$R_{\text{total}}^{(2)} = 2\bigl(G + R_1\bigr) - G = 2G + 2R_1 - G = G + 2R_1$$

Determining the additional resistance. The voltmeter already possesses $$R_1$$ in series. Therefore the extra resistance that must be inserted in series, call it $$R_{\text{add}}$$, is the difference between the new required total and the resistance already present:

$$R_{\text{add}} = R_{\text{total}}^{(2)} - R_1 = \bigl(G + 2R_1\bigr) - R_1 = R_1 + G$$

Thus the additional resistance to be connected in series with the existing $$R_1$$ is $$R_1 + G$$.

Hence, the correct answer is Option D.