Ten identical cells connected is series are needed to heated a wire of length one meter and radius 'r'by `10^(@)C` in time t. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time 't' ?

Options

A. 10
B. 20
C. 30
D. 40

Answer

To solve the problem, we need to understand the relationship between the length of the wire, the resistance, and the amount of heat generated. The heat generated in a conductor can be calculated using the formula:

\[ Q = I^2 R t \]

Where:
- \( Q \) is the heat generated,
- \( I \) is the current flowing through the wire,
- \( R \) is the resistance of the wire,
- \( t \) is the time for which the current flows.

### Step-by-Step Solution:

1. **Determine the Resistance of the Wire**:
The resistance \( R \) of a wire is given by 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 of the wire. The cross-sectional area \( A \) can be calculated as:
\[ A = \pi r^2 \]
For a wire of length 1 meter and radius \( r \):
\[ R_1 = \frac{\rho \cdot 1}{\pi r^2} = \frac{\rho}{\pi r^2} \]

2. **Heat Required for 1 Meter Wire**:
According to the problem, 10 cells are needed to heat a 1-meter wire by \( 10^\circ C \) in time \( t \). Thus, the heat generated is:
\[ Q_1 = I^2 R_1 t \]

3. **Resistance of the 2 Meter Wire**:
For a wire of length 2 meters and the same radius \( r \):
\[ R_2 = \frac{\rho \cdot 2}{\pi r^2} = \frac{2\rho}{\pi r^2} \]

4. **Heat Required for 2 Meter Wire**:
The heat required to raise the temperature of the 2-meter wire by \( 10^\circ C \) in the same time \( t \) is:
\[ Q_2 = I^2 R_2 t = I^2 \left(\frac{2\rho}{\pi r^2}\right) t \]

5. **Relating Heat Generated**:
Since the heat generated is proportional to the resistance, we can set up the ratio of the heat generated for the two wires:
\[ \frac{Q_2}{Q_1} = \frac{R_2}{R_1} = \frac{\frac{2\rho}{\pi r^2}}{\frac{\rho}{\pi r^2}} = 2 \]
This means that the heat required for the 2-meter wire is twice that of the 1-meter wire.

6. **Calculating the Number of Cells**:
Since 10 cells are required for the 1-meter wire, and the heat required for the 2-meter wire is double, we will need:
\[ \text{Number of cells for 2-meter wire} = 10 \times 2 = 20 \]

### Final Answer:
Therefore, **20 cells** will be required to heat the wire of length 2 meters by \( 10^\circ C \) in time \( t \).