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 analyze the given conditions and apply the principles of a meter bridge.

### Step 1: Understand the initial setup
In the meter bridge, we have:
- Resistance \( R_1 = 2 \, \Omega \) in the left gap.
- Unknown resistance \( R_x \) in the right gap.
- The balance point is at \( L_1 = 40 \, \text{cm} \) from the zero end.

### Step 2: Apply the balance condition of the meter bridge
The balance condition for a meter bridge is given by the formula:
\[ \frac{R_1}{R_x} = \frac{L_1}{100 - L_1} \]
Substituting the known values:
\[ \frac{2}{R_x} = \frac{40}{100 - 40} \]
This simplifies to:
\[ \frac{2}{R_x} = \frac{40}{60} \]
\[ \frac{2}{R_x} = \frac{2}{3} \]

### Step 3: Solve for the unknown resistance \( R_x \)
Cross-multiplying gives:
\[ 2 \cdot 3 = 2 \cdot R_x \]
\[ 6 = 2R_x \]
Dividing both sides by 2:
\[ R_x = 3 \, \Omega \]

### Step 4: Analyze the new condition after shunting
Now, we shunt the unknown resistance \( R_x \) with \( 2 \, \Omega \). The new resistance \( R' \) in parallel can be calculated using the formula for resistances in parallel:
\[ \frac{1}{R'} = \frac{1}{R_x} + \frac{1}{2} \]
Substituting \( R_x = 3 \, \Omega \):
\[ \frac{1}{R'} = \frac{1}{3} + \frac{1}{2} \]
Finding a common denominator (which is 6):
\[ \frac{1}{R'} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \]
Thus,
\[ R' = \frac{6}{5} = 1.2 \, \Omega \]

### Step 5: Find the new balance point
Now, we need to find the new balance point \( L_2 \) using the new resistance \( R' \):
\[ \frac{R_1}{R'} = \frac{L_2}{100 - L_2} \]
Substituting the values:
\[ \frac{2}{1.2} = \frac{L_2}{100 - L_2} \]
Cross-multiplying gives:
\[ 2(100 - L_2) = 1.2 L_2 \]
Expanding and rearranging:
\[ 200 - 2L_2 = 1.2L_2 \]
\[ 200 = 3.2L_2 \]
Dividing both sides by 3.2:
\[ L_2 = \frac{200}{3.2} = 62.5 \, \text{cm} \]

### Step 6: Calculate the shift in the balance point
The shift in the balance point is given by:
\[ \text{Shift} = L_2 - L_1 = 62.5 \, \text{cm} - 40 \, \text{cm} = 22.5 \, \text{cm} \]

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

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