Two metallic wires of same material and same length have different diameter. When the wires are connected in parallel across an ideal battery the rate of heat produced in thinner wire is `Q_(1)` and that in thicker wire is `Q_(1)`. The correct statements is

Options

A. `Q_(1) = Q_(2)`
B. `Q_(1) lt Q_(2)`
C. `Q_(1) gt Q_(2)`
D. It will depend on the emf of the battery

Answer

To solve the problem, we need to analyze the situation with the two metallic wires connected in parallel across an ideal battery.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Two metallic wires of the same material and the same length.
- Different diameters (hence different cross-sectional areas).
- Connected in parallel across an ideal battery.

2. **Understand the Concept of Resistance**:
- The resistance \( R \) of a wire can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

3. **Calculate the Cross-Sectional Area**:
- The cross-sectional area \( A \) for a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]
- For a thinner wire (let's call it wire 1) and a thicker wire (wire 2), we denote their areas as \( A_1 \) and \( A_2 \) respectively, where \( A_1 < A_2 \).

4. **Determine the Resistance of Each Wire**:
- For the thinner wire (wire 1):
\[ R_1 = \frac{\rho L}{A_1} \]
- For the thicker wire (wire 2):
\[ R_2 = \frac{\rho L}{A_2} \]
- Since \( A_1 < A_2 \), it follows that \( R_1 > R_2 \).

5. **Power Dissipation in Each Wire**:
- The power \( P \) dissipated in a resistor when connected to a voltage \( V \) is given by:
\[ P = \frac{V^2}{R} \]
- Since both wires are connected in parallel, they experience the same voltage \( V \).

6. **Calculate Power for Each Wire**:
- For the thinner wire (wire 1):
\[ P_1 = \frac{V^2}{R_1} \]
- For the thicker wire (wire 2):
\[ P_2 = \frac{V^2}{R_2} \]
- Since \( R_1 > R_2 \), it follows that \( P_1 < P_2 \).

7. **Relate Power to Heat Produced**:
- The rate of heat produced \( Q \) in a wire is directly proportional to the power dissipated in that wire. Thus:
\[ Q_1 \propto P_1 \quad \text{and} \quad Q_2 \propto P_2 \]
- Therefore, since \( P_1 < P_2 \), we conclude that:
\[ Q_1 < Q_2 \]

### Conclusion:
The rate of heat produced in the thinner wire \( Q_1 \) is less than that in the thicker wire \( Q_2 \).

### Correct Statement:
Thus, the correct statement is \( Q_1 < Q_2 \).

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