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) gt 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,

Options

A. a)`R_(A)R_(V)=G^(2)((I_(0)-I_(g))/(I_(g)))` and `(R_(A))/(R_(V))=((I_(g))/(I_(0)-I_(g)))`
B. b) `R_(A)R_(V)=G^(2)` and `(R_(A))/(R_(V))=((I_(g))/(I_(0)-I_(g)))^(2)`
C. c) `R_(A)R_(V)=G^(2)((I_(g))/(I_(0)-I_(g)))` and `(R_(A))/(R_(V))=((I_(0)-I_(g))/(I_(g)))`
D. d) `R_(A)R_(V)=G^(2)` and `(R_(A))/(R_(V))=(I_(g))/((I_(0)-I_(g)))`

Answer

To solve the problem, we need to derive the relationships for the shunt resistance \( R_A \) when the galvanometer is converted into an ammeter and the series resistance \( R_V \) when it is converted into a voltmeter.

### Step-by-Step Solution:

1. **Understanding the Galvanometer:**
- A moving coil galvanometer has a resistance \( G \) and shows full-scale deflection for a current \( I_g \).

2. **Conversion to Ammeter:**
- To convert the galvanometer into an ammeter, we connect a shunt resistance \( R_A \) in parallel with the galvanometer.
- The total current \( I_0 \) flowing through the circuit is divided between the galvanometer and the shunt:
\[ I_g = I_0 - I_s \]
where \( I_s \) is the current through the shunt.
- The voltage across both the galvanometer and the shunt is the same:
\[ I_g \cdot G = I_s \cdot R_A \]
- Substituting \( I_s = I_0 - I_g \) into the equation gives:
\[ I_g \cdot G = (I_0 - I_g) \cdot R_A \]
- Rearranging for \( R_A \):
\[ R_A = \frac{I_g}{I_0 - I_g} \cdot G \tag{1} \]

3. **Conversion to Voltmeter:**
- To convert the galvanometer into a voltmeter, we connect a series resistance \( R_V \).
- The total voltage across the galvanometer and the series resistance is:
\[ V = I_g \cdot G + I_g \cdot R_V \]
- Since \( V = G \cdot I_0 \) (the maximum voltage for the voltmeter), we have:
\[ G \cdot I_0 = I_g \cdot G + I_g \cdot R_V \]
- Rearranging for \( R_V \):
\[ R_V = \frac{I_0 - I_g}{I_g} \cdot G \tag{2} \]

4. **Finding the Product \( R_A \cdot R_V \):**
- Multiply equations (1) and (2):
\[ R_A \cdot R_V = \left( \frac{I_g}{I_0 - I_g} \cdot G \right) \cdot \left( \frac{I_0 - I_g}{I_g} \cdot G \right) \]
- Simplifying this:
\[ R_A \cdot R_V = G^2 \]

5. **Finding the Ratio \( \frac{R_A}{R_V} \):**
- Dividing equation (1) by equation (2):
\[ \frac{R_A}{R_V} = \frac{\frac{I_g}{I_0 - I_g} \cdot G}{\frac{I_0 - I_g}{I_g} \cdot G} \]
- Simplifying this gives:
\[ \frac{R_A}{R_V} = \left( \frac{I_g}{I_0 - I_g} \right) \cdot \left( \frac{I_g}{I_0 - I_g} \right) = \left( \frac{I_g}{I_0 - I_g} \right)^2 \]

### Final Results:
- \( R_A \cdot R_V = G^2 \)
- \( \frac{R_A}{R_V} = \left( \frac{I_g}{I_0 - I_g} \right)^2 \)

### Conclusion:
The correct option based on the derived equations is option B.