In a meter bridge, the left and right gaps are closed by resistances 2 ohm and 3 ohm respectively. The value of shunt to be connected to 3 ohm resistor to shift the balancing point by 22.5 cm is

Options

A. 3 ohm
B. 1.7 ohm
C. 1 ohm
D. 2 ohm

Answer

To solve the problem step by step, we will follow the principles of the meter bridge and the relationships between resistances and lengths.

### Step-by-Step Solution:

1. **Understanding the Meter Bridge Setup**:
- In a meter bridge, the left gap has a resistance \( R_1 = 2 \, \Omega \) and the right gap has a resistance \( R_2 = 3 \, \Omega \).
- The balancing point is initially at \( L_1 \) cm from the left end.

2. **Finding the Initial Balancing Point**:
- The formula for the meter bridge is given by:
\[ \frac{R_1}{R_2} = \frac{L_1}{100 - L_1} \]
- Substituting the values:
\[ \frac{2}{3} = \frac{L_1}{100 - L_1} \]
- Cross-multiplying gives:
\[ 2(100 - L_1) = 3L_1 \]
- Expanding and rearranging:
\[ 200 - 2L_1 = 3L_1 \implies 200 = 5L_1 \implies L_1 = \frac{200}{5} = 40 \, \text{cm} \]

3. **Introducing the Shunt Resistance**:
- A shunt resistance \( S \) is connected in parallel with the 3 ohm resistor.
- The equivalent resistance \( R_{eq} \) of the 3 ohm resistor and the shunt is given by:
\[ R_{eq} = \frac{3S}{S + 3} \]

4. **Finding the New Balancing Point**:
- The new balancing point \( L_2 \) after shifting is:
\[ L_2 = L_1 + 22.5 = 40 + 22.5 = 62.5 \, \text{cm} \]
- The new length on the right side is:
\[ 100 - L_2 = 100 - 62.5 = 37.5 \, \text{cm} \]

5. **Setting Up the New Balance Equation**:
- The new balance condition is:
\[ \frac{R_1}{R_{eq}} = \frac{L_2}{100 - L_2} \]
- Substituting the values:
\[ \frac{2}{\frac{3S}{S + 3}} = \frac{62.5}{37.5} \]
- Simplifying the right side:
\[ \frac{62.5}{37.5} = \frac{5}{3} \]
- Thus, we have:
\[ \frac{2(S + 3)}{3S} = \frac{5}{3} \]

6. **Cross-Multiplying to Solve for S**:
- Cross-multiplying gives:
\[ 2(S + 3) = 5S \]
- Expanding:
\[ 2S + 6 = 5S \]
- Rearranging:
\[ 6 = 5S - 2S \implies 3S = 6 \implies S = \frac{6}{3} = 2 \, \Omega \]

7. **Conclusion**:
- The value of the shunt resistance \( S \) that needs to be connected to the 3 ohm resistor to shift the balancing point by 22.5 cm is \( 2 \, \Omega \).

### Final Answer:
The value of the shunt resistance is \( 2 \, \Omega \).