In a metre bridge when the resistance in the left gap is `2Omega` and an unknown resistance in the right gap, the balance point is obtained at 40 cm from the zero end. On shunting the unknown resistance with `2Omega` find the shift of the balance point on the bridge wire.

Answer

To solve the problem step by step, we will follow the principles of the meter bridge and the concept of parallel resistances.

### Step 1: Understand the initial setup
In the meter bridge, we have:
- Resistance \( R_1 = 2 \, \Omega \) (left gap)
- Unknown resistance \( R_0 \) (right gap)
- Balance point at \( x = 40 \, \text{cm} \)

### Step 2: Use the balance condition
According to the principle of the meter bridge, the balance point is given by the relation:
\[ \frac{R_1}{R_0} = \frac{x}{100 - x} \]
Substituting the known values:
\[ \frac{2}{R_0} = \frac{40}{100 - 40} \]
This simplifies to:
\[ \frac{2}{R_0} = \frac{40}{60} = \frac{2}{3} \]

### Step 3: Solve for the unknown resistance \( R_0 \)
Cross-multiplying gives:
\[ 2 \cdot 3 = 2 \cdot R_0 \]
\[ R_0 = 3 \, \Omega \]

### Step 4: Introduce the shunt resistance
Now, we shunt the unknown resistance \( R_0 \) with \( 2 \, \Omega \). The equivalent resistance \( R_0' \) can be calculated using the formula for resistances in parallel:
\[ R_0' = \frac{R_0 \cdot 2}{R_0 + 2} \]
Substituting \( R_0 = 3 \, \Omega \):
\[ R_0' = \frac{3 \cdot 2}{3 + 2} = \frac{6}{5} \, \Omega \]

### Step 5: Find the new balance point
Now, we need to find the new balance point \( x' \) using the new resistance \( R_0' \):
\[ \frac{R_1}{R_0'} = \frac{x'}{100 - x'} \]
Substituting the values:
\[ \frac{2}{\frac{6}{5}} = \frac{x'}{100 - x'} \]
This simplifies to:
\[ \frac{2 \cdot 5}{6} = \frac{x'}{100 - x'} \]
\[ \frac{5}{3} = \frac{x'}{100 - x'} \]

### Step 6: Cross-multiply to find \( x' \)
Cross-multiplying gives:
\[ 5(100 - x') = 3x' \]
Expanding this:
\[ 500 - 5x' = 3x' \]
Combining like terms:
\[ 500 = 8x' \]
Thus, we find:
\[ x' = \frac{500}{8} = 62.5 \, \text{cm} \]

### Step 7: Calculate the shift in the balance point
The shift in the balance point is given by:
\[ \text{Shift} = x' - x = 62.5 \, \text{cm} - 40 \, \text{cm} = 22.5 \, \text{cm} \]

### Final Answer
The shift of the balance point on the bridge wire is \( 22.5 \, \text{cm} \).

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