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

Options

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

Explanation

To solve the problem step by step, we need to find the value of the shunt resistance (R) to be added to the 3 ohm resistor in a meter bridge setup, which shifts the balancing point by 22.5 cm.

### Step-by-Step Solution:

**Step 1: Understand the setup of the meter bridge.**
- In a meter bridge, two resistances are connected in the gaps of the bridge. Here, we have R1 = 2 ohms and R2 = 3 ohms.

**Hint:** Remember that the meter bridge operates on the principle of balancing the two sides of the bridge with the resistances.

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**Step 2: Calculate the initial balancing length (L).**
- The balancing condition is given by the formula:
\[ \frac{R1}{R2} = \frac{L}{100 - L} \]
- Substituting R1 = 2 ohms and R2 = 3 ohms:
\[ \frac{2}{3} = \frac{L}{100 - L} \]
- Cross-multiplying gives:
\[ 2(100 - L) = 3L \]
- Expanding and rearranging:
\[ 200 - 2L = 3L \implies 5L = 200 \implies L = 40 \text{ cm} \]

**Hint:** This step shows how to find the initial balancing length using the known resistances.

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**Step 3: Determine the new balancing length after adding the shunt.**
- The problem states that the balancing point shifts by 22.5 cm. Therefore, the new balancing length (L') is:
\[ L' = 40 \text{ cm} + 22.5 \text{ cm} = 62.5 \text{ cm} \]

**Hint:** Make sure to add the shift to the original balancing length to find the new length.

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**Step 4: Set up the equation for the new balancing condition with the shunt.**
- Let R be the shunt resistance added to the 3 ohm resistor. The equivalent resistance (R_eq) of the 3 ohm resistor and the shunt in parallel is given by:
\[ R_{eq} = \frac{3R}{3 + R} \]
- Now, using the new balancing length:
\[ \frac{2}{R_{eq}} = \frac{62.5}{100 - 62.5} = \frac{62.5}{37.5} = \frac{5}{3} \]

**Hint:** Remember to express the balancing condition in terms of the equivalent resistance after adding the shunt.

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**Step 5: Solve for the equivalent resistance.**
- From the balancing condition:
\[ \frac{2}{R_{eq}} = \frac{5}{3} \implies R_{eq} = \frac{2 \cdot 3}{5} = \frac{6}{5} \text{ ohms} \]

**Hint:** Use algebraic manipulation to isolate R_eq.

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**Step 6: Set up the equation for the equivalent resistance.**
- We have:
\[ \frac{3R}{3 + R} = \frac{6}{5} \]
- Cross-multiplying gives:
\[ 5 \cdot 3R = 6(3 + R) \implies 15R = 18 + 6R \]
- Rearranging gives:
\[ 15R - 6R = 18 \implies 9R = 18 \implies R = 2 \text{ ohms} \]

**Hint:** Make sure to carefully rearrange the equation to isolate R.

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

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