When a wire of uniform cross-section `a`, length `I` and resistance `R` is bent into a complete circle, resistance between two of diametrically opposite points will be

Options

A. `(R)/(4)`
B. `(R)/(8)`
C. `4R`
D. `(R)/(2)`

Answer

To solve the problem of finding the resistance between two diametrically opposite points when a wire of uniform cross-section \( a \), length \( L \), and resistance \( R \) is bent into a complete circle, we can follow these steps:

### Step 1: Understand the Configuration
When the wire is bent into a complete circle, it can be visualized as two semicircles. Each semicircle will have half the length of the original wire.

### Step 2: Calculate the Length of Each Semicircle
The total length of the wire is \( L \). When bent into a circle, the length of each semicircle is:
\[ L_{\text{semicircle}} = \frac{L}{2} \]

### Step 3: Determine the Resistance of Each Semicircle
The resistance \( R \) of the wire is given by:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Since the wire is now in two semicircles, the resistance of each semicircle can be calculated as:
\[ R_{\text{semicircle}} = \frac{R}{2} \]

### Step 4: Analyze the Resistance Between Two Opposite Points
The resistance between two diametrically opposite points involves the two semicircles. The resistances of the two semicircles are in parallel. The formula for the equivalent resistance \( R_P \) of two resistances \( R_1 \) and \( R_2 \) in parallel is:
\[ \frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} \]
In this case, both \( R_1 \) and \( R_2 \) are equal to \( \frac{R}{2} \):
\[ \frac{1}{R_P} = \frac{1}{\frac{R}{2}} + \frac{1}{\frac{R}{2}} = \frac{2}{R/2} = \frac{4}{R} \]

### Step 5: Solve for the Equivalent Resistance
Now, we can find \( R_P \):
\[ R_P = \frac{R}{4} \]

### Conclusion
The resistance between two diametrically opposite points when the wire is bent into a complete circle is:
\[ \boxed{\frac{R}{4}} \]

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