A wire of length L and 3 identical cells of negligible internal resistances are connected is series . Due to the current , the temperature of the wire is raised by `DeltaT` in time t.N number of similar cells is now connected in series with 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

Answer

To solve the problem, we will analyze the heat generated in both cases and equate them to find the value of N.

### Step-by-Step Solution:

1. **Understanding the First Case:**
- We have a wire of length \( L \) connected to 3 identical cells.
- The resistance \( R \) of the wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
- The total voltage provided by the cells is \( V = 3\epsilon \).
- The current \( I \) flowing through the wire is given by Ohm's law:
\[ I = \frac{V}{R} = \frac{3\epsilon}{R} \]

2. **Calculating Heat Produced in the First Case:**
- The heat \( H \) produced in time \( t \) is given by:
\[ H = I^2 R t \]
- Substituting the expression for current:
\[ H = \left(\frac{3\epsilon}{R}\right)^2 R t = \frac{9\epsilon^2}{R} t \]

3. **Relating Heat to Temperature Change:**
- The heat produced raises the temperature of the wire by \( \Delta T \). The heat can also be expressed as:
\[ H = m s \Delta T \]
- The mass \( m \) of the wire is given by:
\[ m = \rho \cdot V = \rho \cdot A \cdot L \]
- Therefore, we can write:
\[ H = \rho A L s \Delta T \]

4. **Equating the Two Expressions for Heat:**
- From the two expressions for heat, we have:
\[ \frac{9\epsilon^2}{R} t = \rho A L s \Delta T \]
- Rearranging gives us:
\[ \Delta T = \frac{9\epsilon^2 t}{\rho A L s R} \]

5. **Understanding the Second Case:**
- Now, we connect \( N \) identical cells in series with a wire of length \( 2L \).
- The new resistance \( R' \) is:
\[ R' = \frac{\rho (2L)}{A} = 2R \]
- The total voltage provided by the cells is \( V' = N\epsilon \).
- The current \( I' \) flowing through the wire is:
\[ I' = \frac{N\epsilon}{R'} = \frac{N\epsilon}{2R} \]

6. **Calculating Heat Produced in the Second Case:**
- The heat \( H' \) produced in time \( t \) is:
\[ H' = (I')^2 R' t = \left(\frac{N\epsilon}{2R}\right)^2 (2R) t = \frac{N^2 \epsilon^2}{4R} t \]

7. **Relating Heat to Temperature Change in the Second Case:**
- The heat produced raises the temperature of the wire by \( \Delta T \):
\[ H' = 2m s \Delta T \]
- The mass of the new wire of length \( 2L \) is:
\[ m' = \rho A (2L) = 2m \]
- Thus:
\[ H' = 2(\rho A L) s \Delta T \]

8. **Equating the Two Expressions for Heat in the Second Case:**
- From the two expressions for heat, we have:
\[ \frac{N^2 \epsilon^2}{4R} t = 2(\rho A L) s \Delta T \]
- Rearranging gives us:
\[ \Delta T = \frac{N^2 \epsilon^2 t}{8 \rho A L s R} \]

9. **Equating the Temperature Changes:**
- Since both cases raise the temperature by the same amount \( \Delta T \):
\[ \frac{9\epsilon^2 t}{\rho A L s R} = \frac{N^2 \epsilon^2 t}{8 \rho A L s R} \]
- Canceling common terms:
\[ 9 = \frac{N^2}{8} \]
- Solving for \( N^2 \):
\[ N^2 = 72 \quad \Rightarrow \quad N = \sqrt{72} = 6 \]

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