A uniform wire of resistance `9 Omega` is joined end-to-end to from a circle. Then, the resistance of the circular wire between any two diametrically points is

Options

A. `6 Omega`
B. `3 Omega`
C. `(9)/(4)Omega`
D. `(3)/(2)Omega`

Answer

To find the resistance of a circular wire formed by joining a uniform wire of resistance \(9 \, \Omega\) end-to-end between any two diametrically opposite points, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a circular wire made from a uniform wire with a total resistance of \(9 \, \Omega\). We need to find the resistance between two points that are diametrically opposite on the circle.

2. **Dividing the Circle**:
When we consider two diametrically opposite points on the circle, the circular wire can be thought of as being divided into two equal halves. Each half will have a resistance of:
\[ R_{\text{half}} = \frac{R_{\text{total}}}{2} = \frac{9 \, \Omega}{2} = 4.5 \, \Omega \]

3. **Setting Up the Circuit**:
The two halves of the wire can be treated as two resistors in parallel. Let’s denote the resistance of each half as \(R_1\) and \(R_2\):
\[ R_1 = R_2 = 4.5 \, \Omega \]

4. **Calculating Equivalent Resistance**:
The formula for the equivalent resistance \(R'\) of two resistors in parallel is given by:
\[ \frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the values:
\[ \frac{1}{R'} = \frac{1}{4.5} + \frac{1}{4.5} = \frac{2}{4.5} = \frac{2}{\frac{9}{2}} = \frac{4}{9} \]
Therefore,
\[ R' = \frac{9}{4} \, \Omega \]

5. **Conclusion**:
The resistance of the circular wire between any two diametrically opposite points is:
\[ R' = \frac{9}{4} \, \Omega \]

### Final Answer:
The resistance between any two diametrically opposite points is \( \frac{9}{4} \, \Omega \).