A wire of length L and 3 identical cells of negligible internal resistance are connected in series. Due to the current, the temperature of the wire is raised by `DeltaT` in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount `DeltaT` in the same time t. the value of N is

Options

A. 4
B. 6
C. 8
D. 9

Explanation

To solve the problem, we need to analyze the two scenarios given in the question step by step.

### Step 1: Analyze the first scenario
We have a wire of length \( L \) connected to three identical cells in series. The total EMF (voltage) provided by the cells is:
\[ V_{\text{total}} = 3V \]
where \( V \) is the voltage of one cell.

### Step 2: Calculate the resistance of the wire
The resistance \( R \) of the wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

### Step 3: Calculate the heat generated in the first scenario
The power \( P \) dissipated in the wire can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
In this case, substituting the total EMF:
\[ P_1 = \frac{(3V)^2}{R} = \frac{9V^2}{R} \]
The heat generated \( H_1 \) in time \( t \) is given by:
\[ H_1 = P_1 \cdot t = \frac{9V^2}{R} \cdot t \]

### Step 4: Relate heat to temperature change
From the specific heat formula, the heat generated can also be expressed as:
\[ H_1 = mc\Delta T \]
where \( m \) is the mass of the wire, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.

### Step 5: Express mass in terms of length
The mass \( m \) of the wire can be expressed as:
\[ m = \rho \cdot V = \rho \cdot A \cdot L \]
Thus, we can rewrite the heat equation as:
\[ H_1 = \rho A L c \Delta T \]

### Step 6: Set up the equation for the first scenario
Equating the two expressions for heat generated:
\[ \frac{9V^2}{R} \cdot t = \rho A L c \Delta T \]

### Step 7: Analyze the second scenario
Now, we connect \( N \) similar cells in series with a wire of length \( 2L \). The total EMF in this case is:
\[ V_{\text{total}} = NV \]
The resistance of the new wire of length \( 2L \) is:
\[ R' = \frac{\rho (2L)}{A} = 2R \]

### Step 8: Calculate the heat generated in the second scenario
The power \( P_2 \) in this case is:
\[ P_2 = \frac{(NV)^2}{R'} = \frac{(NV)^2}{2R} \]
The heat generated \( H_2 \) in time \( t \) is:
\[ H_2 = P_2 \cdot t = \frac{(NV)^2}{2R} \cdot t \]

### Step 9: Relate heat to temperature change for the second scenario
For the second wire, the mass \( m' \) is:
\[ m' = \rho A (2L) = 2\rho A L \]
Thus, the heat generated can also be expressed as:
\[ H_2 = 2\rho A L c \Delta T \]

### Step 10: Set up the equation for the second scenario
Equating the two expressions for heat generated:
\[ \frac{(NV)^2}{2R} \cdot t = 2\rho A L c \Delta T \]

### Step 11: Solve for \( N \)
Now we have two equations:
1. \(\frac{9V^2}{R} \cdot t = \rho A L c \Delta T\)
2. \(\frac{(NV)^2}{2R} \cdot t = 2\rho A L c \Delta T\)

Dividing the second equation by the first:
\[ \frac{\frac{(NV)^2}{2R} \cdot t}{\frac{9V^2}{R} \cdot t} = \frac{2\rho A L c \Delta T}{\rho A L c \Delta T} \]
This simplifies to:
\[ \frac{(NV)^2}{2 \cdot 9V^2} = 2 \]
\[ \frac{N^2}{18} = 2 \]
\[ N^2 = 36 \]
\[ N = 6 \]

### Final Answer
The value of \( N \) is \( 6 \).