Two wires `'A'` and `'B'` of the same material have their lengths in the ratio `1 : 2` and radii in the ratio `2 : 1` The two wires are connected in parallel across a battery. The ratio of the heat produced in `'A'` to the heat produced in `'B'` for the same time is

Options

A. `1:2`
B. `2:1`
C. `1:8`
D. `8:1`

Explanation

To solve the problem, we need to find the ratio of the heat produced in wire A to the heat produced in wire B when they are connected in parallel across a battery.

### Step-by-Step Solution:

1. **Identify the Given Ratios:**
- Length of wire A (L1) to length of wire B (L2) is in the ratio 1:2.
- Radius of wire A (R1) to radius of wire B (R2) is in the ratio 2:1.

2. **Calculate the Cross-Sectional Areas:**
- The cross-sectional area (A) of a wire is given by the formula \( A = \pi R^2 \).
- For wire A: \( A_1 = \pi R_1^2 \)
- For wire B: \( A_2 = \pi R_2^2 \)
- Given \( R_1 : R_2 = 2 : 1 \), we can express this as \( R_1 = 2R \) and \( R_2 = R \).
- Therefore, \( A_1 = \pi (2R)^2 = 4\pi R^2 \) and \( A_2 = \pi R^2 \).
- The ratio of areas \( A_1 : A_2 = 4 : 1 \).

3. **Calculate the Resistances:**
- The resistance (R) of a wire is given by \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity.
- For wire A:
\[ R_1 = \frac{\rho L_1}{A_1} = \frac{\rho L_1}{4\pi R^2} \]
- For wire B:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho L_2}{\pi R^2} \]
- Given \( L_1 : L_2 = 1 : 2 \), we can express this as \( L_1 = L \) and \( L_2 = 2L \).
- Thus,
\[ R_1 = \frac{\rho L}{4\pi R^2} \quad \text{and} \quad R_2 = \frac{\rho (2L)}{\pi R^2} = \frac{2\rho L}{\pi R^2} \]

4. **Calculate the Power in Each Wire:**
- The power (P) in a resistor connected to a voltage (V) is given by \( P = \frac{V^2}{R} \).
- For wire A:
\[ P_1 = \frac{V^2}{R_1} = \frac{V^2}{\frac{\rho L}{4\pi R^2}} = \frac{4V^2 \cdot 4\pi R^2}{\rho L} \]
- For wire B:
\[ P_2 = \frac{V^2}{R_2} = \frac{V^2}{\frac{2\rho L}{\pi R^2}} = \frac{V^2 \cdot \pi R^2}{2\rho L} \]

5. **Calculate the Ratio of Powers:**
- The ratio of the powers is:
\[ \frac{P_1}{P_2} = \frac{\frac{4V^2 \cdot 4\pi R^2}{\rho L}}{\frac{V^2 \cdot \pi R^2}{2\rho L}} = \frac{4 \cdot 4}{\frac{1}{2}} = \frac{16}{\frac{1}{2}} = 32 \]

6. **Calculate the Ratio of Heat Produced:**
- Since heat produced is proportional to power when time is constant, the ratio of heat produced in A to heat produced in B is the same as the ratio of their powers:
\[ \frac{H_A}{H_B} = \frac{P_1}{P_2} = 32 \]

### Final Answer:
The ratio of the heat produced in wire A to the heat produced in wire B is \( 8 : 1 \).