A wire of length l is stretched to 4l length keeping volume constant. Then, the new resistance becomes

Options

A. 4 times
B. 2 times
C. 16 times
D. 10 times

Explanation

To solve the problem, we need to determine the new resistance of a wire that has been stretched to four times its original length while keeping its volume constant. Let's break down the steps:

### Step 1: Understand the relationship between resistance, length, and area
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho l}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( l \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Define the initial conditions
Let the initial length of the wire be \( l \) and the initial cross-sectional area be \( A \). The initial resistance \( R_1 \) can be expressed as:
\[ R_1 = \frac{\rho l}{A} \]

### Step 3: Determine the new length and area after stretching
The wire is stretched to a new length \( l' = 4l \). Since the volume of the wire remains constant, we can express the volume before and after stretching:
\[ \text{Initial Volume} = A \cdot l \]
\[ \text{Final Volume} = A' \cdot l' = A' \cdot (4l) \]
Setting these two volumes equal gives us:
\[ A \cdot l = A' \cdot (4l) \]
We can cancel \( l \) from both sides (assuming \( l \neq 0 \)):
\[ A = 4A' \]
This implies:
\[ A' = \frac{A}{4} \]

### Step 4: Calculate the new resistance
Now we can calculate the new resistance \( R_2 \) using the new length and area:
\[ R_2 = \frac{\rho l'}{A'} = \frac{\rho (4l)}{A/4} \]
Substituting \( A' \):
\[ R_2 = \frac{\rho (4l)}{A/4} = \frac{4 \cdot 4 \rho l}{A} = \frac{16 \rho l}{A} \]

### Step 5: Relate the new resistance to the initial resistance
From our initial resistance \( R_1 = \frac{\rho l}{A} \), we can express the new resistance as:
\[ R_2 = 16 \cdot R_1 \]

### Conclusion
Thus, the new resistance becomes 16 times the initial resistance.

\[ \text{New Resistance} = 16 \cdot R_1 \]

### Final Answer
The new resistance becomes **16 times** the initial resistance.

---