A moving coil galvanometer, having a resistance G, produces full scale deflection when a current I$$_G$$ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I$$_0$$ (I$$_0$$ > I$$_g$$) by connecting a shunt resistance R$$_A$$ to it and (ii) into a voltmeter of range 0 to V (V = GI$$_0$$) by connecting a series resistance R$$_V$$ to it. Then,

We begin with a moving-coil galvanometer of resistance $$G$$. Whenever a current $$I_G$$ passes through it, the pointer shows full-scale deflection.

First we convert this galvanometer into an ammeter of range $$0$$ to $$I_0$$, where $$I_0 > I_G$$. For that purpose we connect a shunt resistance $$R_A$$ in parallel with the galvanometer.

In the parallel arrangement the potential difference across the galvanometer equals the potential difference across the shunt. Thus

$$I_G\,G \;=\;(I_0 - I_G)\,R_A\;.$$

Now we solve for $$R_A$$ algebraically. Dividing both sides by $$(I_0 - I_G)$$ and then by $$1$$ (to keep the steps explicit) we get

$$R_A \;=\;\frac{I_G\,G}{I_0 - I_G}\;.$$

Next, we convert the same galvanometer into a voltmeter capable of measuring from $$0$$ to a voltage $$V$$. The question already tells us that this voltage is related to the ammeter range by $$V = G I_0$$. To obtain such a voltmeter we place a series resistance $$R_V$$ with the galvanometer.

The current through the voltmeter at full scale must still be $$I_G$$ (any larger current would damage the movement). Hence Ohm’s law for the whole series combination gives

$$V \;=\; I_G\,(G + R_V)\;.$$

We substitute the given value $$V = G I_0$$ and obtain

$$G I_0 \;=\; I_G\,(G + R_V)\;.$$

Now we solve explicitly for $$R_V$$. First divide both sides by $$I_G$$:

$$\frac{G I_0}{I_G} \;=\; G + R_V\;.$$

Next subtract $$G$$ from both sides:

$$R_V \;=\; \frac{G I_0}{I_G} - G\;.$$

Taking $$G$$ common in the right‐hand expression gives

$$R_V \;=\; G\!\left(\frac{I_0}{I_G} - 1\right)\;=\; G\,\frac{I_0 - I_G}{I_G}\;.$$

We now have explicit expressions for both resistances:

$$R_A = \frac{I_G\,G}{\,I_0 - I_G\,}, \qquad R_V = G\,\frac{I_0 - I_G}{I_G}\;.$$

To see what products and ratios these expressions give, we first multiply the two:

$$R_A\,R_V \;=\; \frac{I_G\,G}{I_0 - I_G} \times G\,\frac{I_0 - I_G}{I_G}\;.$$

The factors $$I_G$$ and $$I_G$$ cancel, as do the factors $$(I_0 - I_G)$$. We are left with

$$R_A\,R_V \;=\; G^2\;.$$

Next, we form the ratio $$\dfrac{R_A}{R_V}$$. Using the same explicit expressions we obtain

$$\frac{R_A}{R_V} \;=\; \frac{\displaystyle\frac{I_G\,G}{I_0 - I_G}} {\displaystyle G\,\frac{I_0 - I_G}{I_G}} \;=\; \frac{I_G\,G}{I_0 - I_G}\times\frac{I_G}{G\,(I_0 - I_G)}\;.$$

The factor $$G$$ cancels from numerator and denominator, giving

$$\frac{R_A}{R_V} \;=\; \frac{I_G^2}{(I_0 - I_G)^2}\;=\;\left(\frac{I_G}{I_0 - I_G}\right)^{\!2}\;.$$

Putting it all together, we have derived

$$R_A\,R_V = G^2,\qquad \frac{R_A}{R_V} = \left(\frac{I_G}{I_0 - I_G}\right)^2\;.$$

These match exactly the relationships stated in Option D.

Hence, the correct answer is Option D.