A conducting wire of uniform cross section, having resistance R is bent to form a circle. O is the centre of this circle. Two points P and Q are selected on the ring in such a manner that `anglePOQ` is `theta`. If a battery of emf E and negligible internal resistance is connected between P and Q, then the current drawn from the battery would be

Options

A. `(E theta)/(4 pi^(2) R (2 pi - theta))`
B. `(E (2 pi - theta))/(4 pi^(2) theta R)`
C. `(4 pi^(2) E theta)/(R(2 pi - theta))`
D. `(4 pi^(2) E)/(theta R(2 pi - theta))`

Explanation

To solve the problem, we need to find the current drawn from a battery connected between two points P and Q on a circular conducting wire of resistance R. The angle subtended by points P and Q at the center O of the circle is θ.

### Step-by-Step Solution:

1. **Understanding the Wire and Resistance**:
- The wire is bent into a circle, and its total resistance is R.
- The circumference of the circle is given by \( C = 2\pi L \), where L is the length of the wire.
- The resistance per unit length of the wire is \( \frac{R}{2\pi L} \).

2. **Finding the Length of Arc PQ**:
- The length of the arc PQ, which subtends an angle θ at the center, is given by:
\[ L_{PQ} = L \cdot \theta \]

3. **Calculating the Resistance R1 between Points P and Q**:
- The resistance \( R_1 \) corresponding to the arc PQ can be calculated as:
\[ R_1 = \text{Resistance per unit length} \times \text{Length of arc PQ} = \frac{R}{2\pi L} \cdot (L \cdot \theta) = \frac{R \theta}{2\pi} \]

4. **Finding the Length of the Remaining Arc**:
- The remaining arc, which is the arc opposite to PQ, subtends an angle \( 2\pi - \theta \) at the center O.
- The length of this arc is:
\[ L_{remaining} = L(2\pi - \theta) \]

5. **Calculating the Resistance R2 of the Remaining Arc**:
- The resistance \( R_2 \) for the remaining arc can be calculated as:
\[ R_2 = \frac{R}{2\pi L} \cdot (L(2\pi - \theta)) = \frac{R(2\pi - \theta)}{2\pi} \]

6. **Finding the Equivalent Resistance (R_eq)**:
- Since \( R_1 \) and \( R_2 \) are in parallel, the equivalent resistance \( R_{eq} \) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
- Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{1}{R_{eq}} = \frac{2\pi}{R\theta} + \frac{2\pi}{R(2\pi - \theta)} \]

7. **Simplifying the Expression**:
- To find \( R_{eq} \), we can simplify:
\[ R_{eq} = \frac{R\theta(2\pi - \theta)}{2\pi} \]

8. **Calculating the Current (I)**:
- Using Ohm's law \( V = IR \), where \( V \) is the EMF \( E \):
\[ I = \frac{E}{R_{eq}} = \frac{E}{\frac{R\theta(2\pi - \theta)}{2\pi}} = \frac{2\pi E}{R\theta(2\pi - \theta)} \]

### Final Answer:
The current drawn from the battery is:
\[ I = \frac{2\pi E}{R\theta(2\pi - \theta)} \]