The same mass of aluminium is draw into two wires 1mm and 2mm thick . Two wires are connected in series and current is passed through them. Heat produced in the wires is in the ratio.

Options

A. `16:1`
B. `8: 32`
C. `8:2`
D. `4:2`

Answer

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

### Step 1: Understand the Heat Produced Formula
The heat produced in a conductor is given by the formula:
\[ H = I^2 R T \]
where:
- \( H \) is the heat produced,
- \( I \) is the current,
- \( R \) is the resistance,
- \( T \) is the time.

### Step 2: Determine the Resistance of Each Wire
The resistance \( R \) of a wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) is the resistivity of the material (constant for both wires since they are made of aluminum),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

For wire 1 (1 mm thick) and wire 2 (2 mm thick), we can write:
- \( R_1 = \frac{\rho L_1}{A_1} \)
- \( R_2 = \frac{\rho L_2}{A_2} \)

### Step 3: Relate the Lengths and Areas of the Wires
Since both wires are made of the same mass of aluminum, and the density is constant, we can express the volume of both wires as equal:
\[ V_1 = V_2 \]
This gives us:
\[ A_1 L_1 = A_2 L_2 \]
From this, we can derive the ratio of lengths:
\[ \frac{L_1}{L_2} = \frac{A_2}{A_1} \]

### Step 4: Write the Heat Produced in Each Wire
Since both wires are in series, the current \( I \) is the same through both wires, and the time \( T \) is also the same. Therefore, we can express the heat produced in each wire as:
- \( H_1 = I^2 R_1 T \)
- \( H_2 = I^2 R_2 T \)

### Step 5: Find the Ratio of Heat Produced
To find the ratio of heat produced in the two wires, we can write:
\[ \frac{H_1}{H_2} = \frac{R_1}{R_2} \]

### Step 6: Substitute the Resistance Values
Substituting the expressions for resistance:
\[ \frac{H_1}{H_2} = \frac{\frac{\rho L_1}{A_1}}{\frac{\rho L_2}{A_2}} = \frac{L_1 A_2}{L_2 A_1} \]

### Step 7: Substitute the Length Ratio
Using the relation from Step 3:
\[ \frac{H_1}{H_2} = \frac{A_2}{A_1} \cdot \frac{A_2}{A_1} = \left(\frac{A_2}{A_1}\right)^2 \]

### Step 8: Calculate the Areas
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
For wire 1 (diameter 1 mm, radius 0.5 mm):
\[ A_1 = \pi (0.5)^2 = \frac{\pi}{4} \text{ mm}^2 \]

For wire 2 (diameter 2 mm, radius 1 mm):
\[ A_2 = \pi (1)^2 = \pi \text{ mm}^2 \]

### Step 9: Find the Ratio of Areas
Now, calculate the ratio:
\[ \frac{A_2}{A_1} = \frac{\pi}{\frac{\pi}{4}} = 4 \]

### Step 10: Substitute Back to Find Heat Ratio
Substituting this back into the heat ratio:
\[ \frac{H_1}{H_2} = 4^2 = 16 \]

### Final Answer
Thus, the ratio of heat produced in the two wires is:
\[ H_1 : H_2 = 16 : 1 \]