A wire has resistance `12 Omega`. It is bent in the form of a circle. The effective resistance between two points across a diameter is.

Options

A. `3 Omega`
B. `6 Omega`
C. `12 Omega`
D. `24 Omega`

Explanation

To find the effective resistance between two points across a diameter of a wire bent in the form of a circle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Resistance of the Wire:**
The total resistance of the wire is given as \( R = 12 \, \Omega \).

2. **Understand the Configuration:**
When the wire is bent into a circle, and we want to find the resistance between two points across a diameter, we can visualize the circle as being divided into two equal halves. Each half will have its own resistance.

3. **Calculate the Resistance of Each Half:**
Since the total resistance of the wire is \( 12 \, \Omega \), when bent into a circle, the resistance of each half (upper and lower) will be:
\[ R_{\text{half}} = \frac{R_{\text{total}}}{2} = \frac{12 \, \Omega}{2} = 6 \, \Omega \]

4. **Model the Resistance Between Points A and B:**
Let’s label the two points across the diameter as point A and point B. The resistance from A to B can be modeled as two resistors (each \( 6 \, \Omega \)) in parallel.

5. **Calculate the Equivalent Resistance:**
For two resistors in parallel, the equivalent resistance \( R_{AB} \) can be calculated using the formula:
\[ \frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting \( R_1 = 6 \, \Omega \) and \( R_2 = 6 \, \Omega \):
\[ \frac{1}{R_{AB}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]
Therefore,
\[ R_{AB} = 3 \, \Omega \]

6. **Conclusion:**
The effective resistance between the two points across the diameter is \( 3 \, \Omega \).

### Final Answer:
The effective resistance between two points across a diameter is \( 3 \, \Omega \).