Consider a wire of length l , area of cross-section A and resistivity `rho` and resistance `(1)/(5) Omega `. Its length is increased by applying a force on I and its length increases by four times of its original length. Find the new resistance in ohms of the wire.

Answer

To find the new resistance of the wire after its length has been increased, we can follow these steps:

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

### Step 2: Identify the initial conditions
We are given that the initial resistance \( R \) is:
\[ R = \frac{1}{5} \, \Omega \]
Let the initial length of the wire be \( L \) and the initial area of cross-section be \( A \).

### Step 3: Determine the new length of the wire
The wire's length is increased to four times its original length:
\[ L' = 4L \]

### Step 4: Use the conservation of volume to find the new area
The volume of the wire remains constant before and after stretching. The volume \( V \) can be expressed as:
\[ V = L \cdot A = L' \cdot A' \]
Substituting \( L' = 4L \):
\[ L \cdot A = 4L \cdot A' \]
Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A = 4A' \]
This implies:
\[ A' = \frac{A}{4} \]

### Step 5: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length and new area:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' = 4L \) and \( A' = \frac{A}{4} \):
\[ R' = \frac{\rho (4L)}{\frac{A}{4}} = \frac{16\rho L}{A} \]

### Step 6: Substitute the initial resistance value
Since we know that \( R = \frac{\rho L}{A} = \frac{1}{5} \):
\[ R' = 16 \cdot \frac{\rho L}{A} = 16 \cdot \frac{1}{5} = \frac{16}{5} \, \Omega \]

### Step 7: Final answer
Thus, the new resistance of the wire is:
\[ R' = \frac{16}{5} \, \Omega \approx 3.2 \, \Omega \]