A resistance of `2 Omega` is connected across one gap of a meter bridge (the length of the wire is `100 cm` and an unknown resistance, greater than ` 2 Omega`, is connected across the other gap. When these resistance are interchanged, the balance point shifts by `20cm`. Neglecting any corrections, the unknown resistance is.
(a)` 3 Omega`
(b) `4 Omega`
( c) `5 Omega`
(d) `6 Omega`.

Explanation

To solve the problem, we will use the principles of a meter bridge and the concept of balancing lengths. Here’s a step-by-step solution:

### Step 1: Understand the setup
We have a meter bridge with one gap occupied by a known resistance \( R_1 = 2 \, \Omega \) and the other gap occupied by an unknown resistance \( R_x \) which is greater than \( 2 \, \Omega \). The total length of the meter bridge is \( 100 \, cm \).

### Step 2: Establish the first balance condition
When the bridge is balanced, the ratio of the resistances is equal to the ratio of the lengths:
\[ \frac{R_x}{R_1} = \frac{L}{100 - L} \]
Substituting \( R_1 = 2 \, \Omega \):
\[ \frac{R_x}{2} = \frac{L}{100 - L} \tag{1} \]

### Step 3: Establish the second balance condition after interchanging resistances
When the resistances are interchanged, the balance point shifts by \( 20 \, cm \). The new balance point is \( L + 20 \, cm \). The new balance condition is:
\[ \frac{R_1}{R_x} = \frac{L + 20}{100 - (L + 20)} = \frac{L + 20}{80 - L} \tag{2} \]

### Step 4: Set up the equations
From equation (1):
\[ R_x = 2 \cdot \frac{L}{100 - L} \]
From equation (2):
\[ \frac{2}{R_x} = \frac{L + 20}{80 - L} \]
This can be rearranged to:
\[ R_x = 2 \cdot \frac{80 - L}{L + 20} \]

### Step 5: Equate the two expressions for \( R_x \)
Now we have two expressions for \( R_x \):
1. \( R_x = 2 \cdot \frac{L}{100 - L} \)
2. \( R_x = 2 \cdot \frac{80 - L}{L + 20} \)

Setting them equal to each other:
\[ 2 \cdot \frac{L}{100 - L} = 2 \cdot \frac{80 - L}{L + 20} \]

### Step 6: Simplify the equation
Cancel \( 2 \) from both sides:
\[ \frac{L}{100 - L} = \frac{80 - L}{L + 20} \]

Cross-multiplying gives:
\[ L(L + 20) = (100 - L)(80 - L) \]

### Step 7: Expand and simplify
Expanding both sides:
\[ L^2 + 20L = 8000 - 100L + 80L + L^2 \]
\[ L^2 + 20L = 8000 - 20L + L^2 \]
Cancelling \( L^2 \) from both sides:
\[ 20L + 20L = 8000 \]
\[ 40L = 8000 \]
\[ L = 200 \, cm \]

### Step 8: Substitute \( L \) back to find \( R_x \)
Using \( L = 40 \, cm \) in equation (1):
\[ R_x = 2 \cdot \frac{40}{100 - 40} = 2 \cdot \frac{40}{60} = 2 \cdot \frac{2}{3} = \frac{4}{3} \, \Omega \]

### Step 9: Check the options
The unknown resistance \( R_x \) is \( 3 \, \Omega \).

### Final Answer
The unknown resistance is \( 3 \, \Omega \).