When a high resistance `'R'` is connected in series with a volmeter of resistance `'G'`, the range of the volmeter increases `5`times. Then `G : R` will be

Options

A. `4 :1`
B. `1 : 2`
C. `8 : 1`
D. `1 : 4`

Explanation

To solve the problem, we need to analyze the situation where a high resistance \( R \) is connected in series with a voltmeter (galvanometer) of resistance \( G \), which increases the range of the voltmeter by 5 times. We will derive the relationship between \( G \) and \( R \).

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Initially, the voltmeter (galvanometer) has a resistance \( G \).
- The voltage across the galvanometer when a current \( I \) flows through it is given by:
\[ V_1 = I \times G \]
- This is our first equation.

2. **Condition with Series Resistance**:
- When a resistance \( R \) is connected in series with the galvanometer, the total resistance becomes \( G + R \).
- The voltage across the new arrangement (voltmeter with series resistance) is:
\[ V_2 = I \times (G + R) \]
- This is our second equation.

3. **Using the Given Information**:
- We know that the new voltage \( V_2 \) is 5 times the original voltage \( V_1 \):
\[ V_2 = 5 \times V_1 \]
- Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ I \times (G + R) = 5 \times (I \times G) \]

4. **Simplifying the Equation**:
- We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ G + R = 5G \]
- Rearranging gives:
\[ R = 5G - G \]
\[ R = 4G \]

5. **Finding the Ratio \( G : R \)**:
- Now we can express the ratio \( \frac{G}{R} \):
\[ \frac{G}{R} = \frac{G}{4G} = \frac{1}{4} \]

6. **Final Result**:
- Therefore, the ratio \( G : R \) is:
\[ G : R = 1 : 4 \]