Same mass of copper is drawn into 2 wires of 1 mm thick and 3mm thick. Two wires are connected in series and current is passed. Heat produced in the wires is the ratio of

Options

A. ` 3:1 `
B. `9:1 `
C. `81:1 `
D. `1: 81`

Answer

To solve the problem of finding the ratio of heat produced in two copper wires of different thicknesses when connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two copper wires of the same mass but different thicknesses: one wire has a thickness of 1 mm and the other has a thickness of 3 mm. We need to find the ratio of heat produced in these wires when a current is passed through them in series.

2. **Volume of Wires**:
Since both wires have the same mass and are made of the same material (copper), they will have the same volume. The volume \( V \) of a wire can be expressed as:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the wire.

3. **Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]
For the first wire (1 mm thick):
\[ A_1 = \frac{\pi (1 \text{ mm})^2}{4} = \frac{\pi}{4} \text{ mm}^2 \]
For the second wire (3 mm thick):
\[ A_2 = \frac{\pi (3 \text{ mm})^2}{4} = \frac{9\pi}{4} \text{ mm}^2 \]

4. **Volume Equivalence**:
Since the volumes of both wires are equal:
\[ A_1 L_1 = A_2 L_2 \]
Substituting the areas:
\[ \frac{\pi}{4} L_1 = \frac{9\pi}{4} L_2 \]
Simplifying gives:
\[ L_1 = 9 L_2 \]

5. **Resistance Calculation**:
The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of copper. For the first wire:
\[ R_1 = \frac{\rho L_1}{A_1} = \frac{\rho (9 L_2)}{\frac{\pi}{4}} = \frac{36 \rho L_2}{\pi} \]
For the second wire:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho L_2}{\frac{9\pi}{4}} = \frac{4 \rho L_2}{9\pi} \]

6. **Heat Produced**:
The heat produced in a resistor is given by:
\[ H = I^2 R \]
Therefore, the heat produced in the first wire \( H_1 \) and the second wire \( H_2 \) can be expressed as:
\[ H_1 = I^2 R_1 = I^2 \cdot \frac{36 \rho L_2}{\pi} \]
\[ H_2 = I^2 R_2 = I^2 \cdot \frac{4 \rho L_2}{9\pi} \]

7. **Finding the Ratio**:
The ratio of heat produced in the two wires is:
\[ \frac{H_1}{H_2} = \frac{I^2 \cdot \frac{36 \rho L_2}{\pi}}{I^2 \cdot \frac{4 \rho L_2}{9\pi}} = \frac{36}{\frac{4}{9}} = 36 \cdot \frac{9}{4} = 81 \]

### Final Answer:
The ratio of heat produced in the wires is \( 81:1 \).