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`

Answer

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. **Understanding the Problem**:
- We have two wires A and B made of the same material.
- The lengths of the wires are in the ratio \( L_A : L_B = 1 : 2 \).
- The radii of the wires are in the ratio \( r_A : r_B = 2 : 1 \).

2. **Finding the Cross-sectional Areas**:
- The area of cross-section \( A \) of a wire is given by the formula \( A = \pi r^2 \).
- Therefore, the area of wire A is:
\[ A_A = \pi r_A^2 \]
- And the area of wire B is:
\[ A_B = \pi r_B^2 \]
- Since \( r_A : r_B = 2 : 1 \), we can express the areas as:
\[ A_A : A_B = \pi (2r)^2 : \pi (r)^2 = 4 : 1 \]

3. **Finding the Resistances**:
- The resistance \( R \) of a wire is given by:
\[ R = \rho \frac{L}{A} \]
- Since both wires are made of the same material, their resistivities \( \rho \) are equal.
- Therefore, the resistance of wire A is:
\[ R_A = \rho \frac{L_A}{A_A} \]
- And the resistance of wire B is:
\[ R_B = \rho \frac{L_B}{A_B} \]
- Substituting the ratios:
\[ R_A : R_B = \frac{L_A / A_A}{L_B / A_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} \]
- Given \( L_A : L_B = 1 : 2 \) and \( A_A : A_B = 4 : 1 \):
\[ R_A : R_B = \frac{1 \cdot 1}{2 \cdot 4} = \frac{1}{8} \]
- Thus, \( R_B : R_A = 8 : 1 \).

4. **Finding the Heat Produced**:
- The heat produced in a resistor when connected to a voltage \( V \) is given by:
\[ H = \frac{V^2}{R} \cdot t \]
- Since the wires are connected in parallel, the voltage across both wires is the same.
- Therefore, the heat produced in wire A is:
\[ H_A = \frac{V^2}{R_A} \cdot t \]
- And for wire B:
\[ H_B = \frac{V^2}{R_B} \cdot t \]
- The ratio of heat produced in A to heat produced in B is:
\[ \frac{H_A}{H_B} = \frac{R_B}{R_A} \]
- Substituting the resistance ratio:
\[ \frac{H_A}{H_B} = \frac{8}{1} = 8 \]

5. **Final Result**:
- Therefore, the ratio of the heat produced in wire A to the heat produced in wire B is:
\[ H_A : H_B = 8 : 1 \]

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