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Introduction:
When a wire is bent in the form of a circle, the effective resistance between the two points on any diameter of the circle can be calculated using the concept of parallel and series combination of resistors.

Solution:
To solve this problem, we can break down the wire into smaller segments, each having a resistance of 12 ohms.

1. Identifying the segments:
Let's assume the wire is divided into 'n' equal segments. Each segment will have a resistance of 12 ohms.

2. Equivalent resistance of two segments:
When two segments are connected in parallel, the total resistance can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Since both resistors are the same, the equation becomes:
\[ \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{12} \]
Simplifying the equation, we get:
\[ \frac{1}{R_{eq}} = \frac{2}{12} \]
\[ \frac{1}{R_{eq}} = \frac{1}{6} \]
Therefore, the equivalent resistance of two segments is 6 ohms.

3. Equivalent resistance of 'n' segments:
To find the equivalent resistance of 'n' segments, we can use the formula for resistors connected in series:
\[ R_{eq} = R_1 + R_2 + R_3 + ... + R_n \]
Substituting the value of resistance for each segment, we get:
\[ R_{eq} = 6 + 6 + 6 + ... + 6 \]
\[ R_{eq} = 6n \]

4. Equivalent resistance of the entire circle:
Since the wire forms a complete circle, there are 'n' segments in total. Therefore, the equivalent resistance of the entire circle is given by:
\[ R_{eq} = 6n \]

5. Relationship between 'n' and the diameter:
The number of segments 'n' can be related to the diameter of the circle. If the diameter of the circle is 'd', then each segment will have a length of 'd/n'. As 'n' approaches infinity, the wire becomes a continuous circle, and the segments become infinitesimally small.

6. Limiting case:
In the limiting case, when 'n' approaches infinity, the equivalent resistance approaches the resistance of a continuous circle. The resistance of a continuous circle is equal to the resistance of a wire bent into a semicircle. The resistance of the semicircle can be calculated using the formula:
\[ R_{eq} = 2R \]
Substituting the value of resistance for each segment, we get:
\[ R_{eq} = 2 \times 12 \]
\[ R_{eq} = 24 \]

Conclusion:
Therefore, the effective resistance between the two points on any diameter of the circle is 24 ohms.

According to my calculation answer is 3ohm