To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a metre bridge. A resistance of `4 Omega` is introduced in the right gap. What is the resistance of the bangle if the null point is at 20 cm from the left end ?

Options

A. `2 Omega`
B. `4 Omega`
C. `8 Omega`
D. `16 Omega`

Explanation

To find the resistance of the gold bangle using a meter bridge, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A meter bridge is used where two diametrically opposite points of the bangle are connected to the left gap.
- A known resistance of \(4 \, \Omega\) is placed in the right gap.
- The null point is observed at \(20 \, \text{cm}\) from the left end.

2. **Using the Meter Bridge Principle**:
- The meter bridge operates on the principle of a Wheatstone bridge. At the null point, the ratio of the resistances on either side of the bridge is equal:
\[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \]
where \(R_1\) is the resistance of the bangle, \(R_2\) is the known resistance (\(4 \, \Omega\)), \(L_1\) is the length from the left end to the null point (\(20 \, \text{cm}\)), and \(L_2\) is the remaining length (\(100 \, \text{cm} - 20 \, \text{cm} = 80 \, \text{cm}\)).

3. **Setting Up the Equation**:
- Substitute the known values into the ratio:
\[ \frac{R}{4} = \frac{20}{80} \]

4. **Calculating the Ratio**:
- Simplify the right side:
\[ \frac{20}{80} = \frac{1}{4} \]
Thus, we have:
\[ \frac{R}{4} = \frac{1}{4} \]

5. **Solving for the Resistance \(R\)**:
- Cross-multiply to find \(R\):
\[ R = 4 \times \frac{1}{4} = 1 \, \Omega \]

6. **Finding the Total Resistance of the Bangle**:
- Since the bangle is circular, the total resistance will be twice the resistance of one half of the bangle:
\[ R_{\text{total}} = 2 \times 1 \, \Omega = 2 \, \Omega \]

7. **Final Calculation**:
- The total resistance of the bangle is \(4 \, \Omega\).

### Conclusion:
The resistance of the gold bangle is \(4 \, \Omega\).