In a meter bridge experment, the value of unknown resistance is `2Omega`. To get tehbalancing point at 40 cm distance form the same end, the resistance in the resistance box will be :

Options

A. 0.5 Omega
B. 3Omega
C. 20 Omega
D. 80 Omega

Explanation

To solve the problem, we need to apply the principle of the meter bridge, which is based on the Wheatstone bridge concept. The balancing condition for a meter bridge is given by the formula:

\[ \frac{R_1}{R_2} = \frac{L}{100 - L} \]

Where:
- \(R_1\) is the known resistance (in this case, the unknown resistance of \(2 \Omega\)),
- \(R_2\) is the resistance in the resistance box (which we need to find),
- \(L\) is the balancing length (given as \(40 \, \text{cm}\)),
- \(100 - L\) is the remaining length of the meter bridge.

### Step 1: Identify the known values
- \(R_1 = 2 \, \Omega\) (the unknown resistance)
- \(L = 40 \, \text{cm}\)

### Step 2: Calculate \(100 - L\)
\[ 100 - L = 100 - 40 = 60 \, \text{cm} \]

### Step 3: Substitute the values into the balancing condition formula
\[ \frac{2 \, \Omega}{R_2} = \frac{40 \, \text{cm}}{60 \, \text{cm}} \]

### Step 4: Simplify the right side of the equation
\[ \frac{40}{60} = \frac{2}{3} \]

### Step 5: Set up the equation
\[ \frac{2}{R_2} = \frac{2}{3} \]

### Step 6: Cross-multiply to solve for \(R_2\)
\[ 2 \cdot 3 = 2 \cdot R_2 \]
\[ 6 = 2R_2 \]

### Step 7: Divide both sides by 2 to find \(R_2\)
\[ R_2 = \frac{6}{2} = 3 \, \Omega \]

### Final Answer
The resistance in the resistance box is \(3 \, \Omega\).

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