A flat circular coil of `n` turns, area `A` and resitance `R` is placed in a uniform magnetic field `B`. The pane of coil is initially perpedicular of `B`. when the coil is rotated through an angle of `180^@` about one of its diameter, a charge `Q_1` flows through the coil When the same coil after charge `Q_2` flows through it. then `Q_2//Q_1` is

Options

A. (a) 1
B. (b) 2
C. (c) `1//2`
D. (d) 0

Explanation

To solve the problem, we will use the principles of electromagnetic induction and the formula for induced electromotive force (emf). The charge flowing through the coil can be calculated using the relationship between charge, current, and time.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The coil has `n` turns, area `A`, and resistance `R`.
- It is placed in a uniform magnetic field `B`, initially perpendicular to the plane of the coil.

2. **Induced EMF Calculation**:
- The magnetic flux (Φ) through the coil is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
- Initially, when the coil is perpendicular to the magnetic field, \(\theta = 0^\circ\) and \(\cos(0) = 1\):
\[ \Phi_1 = B \cdot A \]

3. **Rotation of the Coil**:
- When the coil is rotated through an angle of \(180^\circ\), the new angle \(\theta = 180^\circ\) and \(\cos(180) = -1\):
\[ \Phi_2 = B \cdot A \cdot \cos(180^\circ) = -B \cdot A \]

4. **Change in Magnetic Flux**:
- The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_2 - \Phi_1 = -B \cdot A - (B \cdot A) = -2B \cdot A \]

5. **Induced EMF (ε)**:
- According to Faraday's law of electromagnetic induction, the induced emf (ε) is given by:
\[ \varepsilon = -n \frac{\Delta \Phi}{\Delta t} \]
- Assuming the rotation takes a time Δt, we have:
\[ \varepsilon = -n \frac{-2B \cdot A}{\Delta t} = \frac{2nBA}{\Delta t} \]

6. **Current Calculation**:
- The induced current (I) in the coil can be calculated using Ohm's law:
\[ I = \frac{\varepsilon}{R} = \frac{2nBA}{R \Delta t} \]

7. **Charge Calculation**:
- The charge (Q) that flows through the coil during the time Δt is given by:
\[ Q = I \cdot \Delta t = \left(\frac{2nBA}{R \Delta t}\right) \cdot \Delta t = \frac{2nBA}{R} \]
- Thus, the charge \(Q_1\) when the coil is rotated through \(180^\circ\) is:
\[ Q_1 = \frac{2nBA}{R} \]

8. **Returning to the Original Position**:
- If the coil is rotated back to its original position (0 degrees), the same change in flux occurs, and thus the charge \(Q_2\) will also be:
\[ Q_2 = \frac{2nBA}{R} \]

9. **Finding the Ratio \(Q_2/Q_1\)**:
- Since both \(Q_1\) and \(Q_2\) are equal:
\[ \frac{Q_2}{Q_1} = \frac{\frac{2nBA}{R}}{\frac{2nBA}{R}} = 1 \]

### Final Answer:
\[ \frac{Q_2}{Q_1} = 1 \]