A uniform wire of resistance `4Omega` is bent into circle of radius `r`. As specimen of the same wire is connected along the dimeter of the circle. What is the equivalent resistance across the ends of this wire?

Options

A. `4/((4+pi))Omega`
B. `3/((3+pi)) Omega`
C. `2/((2+pi)Omega`
D. `1/((1+pi)) Omega`

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant formulas.

### Step 1: Understand the configuration
We have a uniform wire with a total resistance of \( R = 4 \, \Omega \). This wire is bent into a circle of radius \( r \). We also have a specimen of the same wire connected along the diameter of the circle.

### Step 2: Determine the length of the wire
The total length of the wire when bent into a circle is given by the circumference of the circle:
\[ L = 2\pi r \]
Since the resistance of the entire wire is \( 4 \, \Omega \), we can find the resistance per unit length of the wire:
\[ \text{Resistance per unit length} = \frac{R}{L} = \frac{4 \, \Omega}{2\pi r} = \frac{2}{\pi r} \, \Omega/m \]

### Step 3: Calculate the resistance of the semicircle
When the wire is bent into a circle, the wire can be considered as two semicircles when we connect the wire along the diameter. The resistance of each semicircle is half of the total resistance of the circle:
\[ R_{\text{semicircle}} = \frac{R}{2} = \frac{4 \, \Omega}{2} = 2 \, \Omega \]

### Step 4: Calculate the resistance of the diameter
The length of the wire along the diameter is equal to the diameter of the circle:
\[ \text{Length along diameter} = 2r \]
Now, we can find the resistance of the wire along the diameter:
\[ R_{\text{diameter}} = \text{Resistance per unit length} \times \text{Length along diameter} = \left(\frac{2}{\pi r}\right) \times (2r) = \frac{4}{\pi} \, \Omega \]

### Step 5: Combine resistances
Now we have three resistances to consider:
1. Resistance of the left semicircle: \( R_1 = 2 \, \Omega \)
2. Resistance of the right semicircle: \( R_2 = 2 \, \Omega \)
3. Resistance of the wire along the diameter: \( R_3 = \frac{4}{\pi} \, \Omega \)

These resistances are in parallel. The formula for the equivalent resistance \( R_{AB} \) for resistors in parallel is given by:
\[ \frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Substituting the values:
\[ \frac{1}{R_{AB}} = \frac{1}{2} + \frac{1}{2} + \frac{\pi}{4} \]
\[ \frac{1}{R_{AB}} = 1 + \frac{\pi}{4} \]
\[ R_{AB} = \frac{1}{1 + \frac{\pi}{4}} = \frac{4}{4 + \pi} \, \Omega \]

### Final Answer
The equivalent resistance across the ends of the wire is:
\[ R_{AB} = \frac{4}{4 + \pi} \, \Omega \]