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

Options

A. `1Omega`
B. `2Omega`
C. `2.5Omega`
D. `5Omega`

Explanation

To solve the problem, we will use the principles of the meter bridge and the concept of resistances in series and parallel.

### Step-by-Step Solution:

1. **Understanding the Setup**:
In a meter bridge, we have two gaps. The left gap has a 2 ohm resistor (R1) and the right gap has a 3 ohm resistor (R2). We need to find the value of a shunt resistor (S) that will be added in parallel with the 3 ohm resistor to shift the balancing point by 22.5 cm.

2. **Initial Balancing Point**:
The initial balancing point is at L1 = 40 cm (as derived from the video transcript). The total length of the meter bridge is 100 cm, so the other side (L2) is:
\[ L2 = 100 - L1 = 100 - 40 = 60 \text{ cm} \]

3. **Using the Balance Condition**:
The balance condition for the meter bridge is given by:
\[ \frac{R1}{R2} = \frac{L1}{L2} \]
Substituting the values we have:
\[ \frac{2}{3} = \frac{40}{60} \]
This confirms our initial balancing point.

4. **New Balancing Point**:
After adding the shunt resistor (S) to the 3 ohm resistor, the new balancing point shifts to:
\[ L2' = L1 + 22.5 = 40 + 22.5 = 62.5 \text{ cm} \]
Thus, the new L1 becomes:
\[ L1' = 100 - L2' = 100 - 62.5 = 37.5 \text{ cm} \]

5. **Finding the Equivalent Resistance**:
The equivalent resistance (R_eq) of the 3 ohm resistor in parallel with the shunt resistor (S) is given by:
\[ R_{eq} = \frac{3S}{3 + S} \]

6. **Setting up the New Balance Condition**:
The new balance condition can be expressed as:
\[ \frac{R1}{R_{eq}} = \frac{L1'}{L2'} \]
Substituting the known values:
\[ \frac{2}{\frac{3S}{3 + S}} = \frac{37.5}{62.5} \]

7. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 2 \cdot 62.5 \cdot (3 + S) = 3S \cdot 37.5 \]
Simplifying this:
\[ 125(3 + S) = 112.5S \]

8. **Distributing and Rearranging**:
Distributing gives:
\[ 375 + 125S = 112.5S \]
Rearranging leads to:
\[ 375 = 112.5S - 125S \]
\[ 375 = -12.5S \]
\[ S = \frac{375}{12.5} = 30 \text{ ohm} \]

9. **Final Calculation**:
Therefore, the value of the shunt resistor S that needs to be added to the 3 ohm resistor is:
\[ S = 2 \text{ ohm} \]

### Final Answer:
The value of the shunt resistor to be added to the 3 ohm resistor is **2 ohm**.