A resistance of `1980Omega` is connected in series with a voltmeter, after which the scale division becomes 100 times larger. Find the resistance of voltmeter.

Options

A. `10Omega`
B. `20Omega`
C. `30Omega`
D. `40Omega`

Answer

To solve the problem, we need to find the resistance of the voltmeter when a resistance of 1980 Ω is connected in series with it, causing the scale division to become 100 times larger.

### Step-by-Step Solution:

1. **Understanding the Initial Setup:**
- Let \( R_v \) be the resistance of the voltmeter.
- Initially, the voltage across each division of the voltmeter is given by \( V = I_G \cdot R_v / n \), where \( I_G \) is the current through the voltmeter and \( n \) is the number of divisions.

2. **Connecting the Resistance in Series:**
- When a resistance of 1980 Ω is connected in series with the voltmeter, the total resistance becomes \( R_v + 1980 \).
- The new voltage across each division now becomes \( V' = I_G \cdot (R_v + 1980) / n \).

3. **Relating the Two Voltages:**
- According to the problem, the new voltage across each division is 100 times the initial voltage:
\[ V' = 100 \cdot V \]
- Substituting the expressions for \( V \) and \( V' \):
\[ \frac{I_G \cdot (R_v + 1980)}{n} = 100 \cdot \frac{I_G \cdot R_v}{n} \]
- We can cancel \( I_G/n \) from both sides (assuming \( I_G \) is not zero):
\[ R_v + 1980 = 100 \cdot R_v \]

4. **Rearranging the Equation:**
- Rearranging the equation gives:
\[ 1980 = 100 R_v - R_v \]
\[ 1980 = 99 R_v \]

5. **Solving for \( R_v \):**
- Dividing both sides by 99:
\[ R_v = \frac{1980}{99} = 20 \, \Omega \]

### Final Answer:
The resistance of the voltmeter \( R_v \) is **20 Ω**.