The resistance of a 10 m long wire is 10 `Omega`. Its length is increased by 25% by stretching the wire uniformly . The resistance of wire will change to

Options

A. `12.5Omega`
B. `14.5Omega`
C. `15.6Omega`
D. `16.6Omega`

Explanation

To solve the problem step by step, we will follow the concepts of resistance, length, and volume of the wire.

### Step 1: Understand the initial conditions
The initial resistance \( R_0 \) of the wire is given as \( 10 \, \Omega \) and its length \( L_0 \) is \( 10 \, m \).

### Step 2: Calculate the new length after stretching
The length of the wire is increased by \( 25\% \). Therefore, the new length \( L' \) can be calculated as:
\[ L' = L_0 + 0.25 \times L_0 = 1.25 \times L_0 \]
Substituting the value of \( L_0 \):
\[ L' = 1.25 \times 10 \, m = 12.5 \, m \]

### Step 3: 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 \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.

### Step 4: Consider volume conservation
When the wire is stretched, its volume remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = A \times L \]
Since the volume before stretching is equal to the volume after stretching, we have:
\[ A_0 \times L_0 = A' \times L' \]
where \( A_0 \) is the original cross-sectional area and \( A' \) is the new cross-sectional area.

### Step 5: Express the new area in terms of the original area
From the volume conservation equation, we can express \( A' \):
\[ A' = \frac{A_0 \times L_0}{L'} \]
Substituting \( L' \):
\[ A' = \frac{A_0 \times L_0}{1.25 \times L_0} = \frac{A_0}{1.25} = 0.8 A_0 \]
This means the new area is \( 80\% \) of the original area.

### Step 6: Calculate the new resistance
Using the resistance formula for the new conditions:
\[ R' = \frac{\rho L'}{A'} \]
Substituting \( L' \) and \( A' \):
\[ R' = \frac{\rho (1.25 L_0)}{0.8 A_0} \]
Now, we can express \( R' \) in terms of the original resistance \( R_0 \):
\[ R' = \frac{1.25}{0.8} \cdot \frac{\rho L_0}{A_0} = \frac{1.25}{0.8} R_0 \]
Substituting \( R_0 = 10 \, \Omega \):
\[ R' = \frac{1.25}{0.8} \times 10 = 1.5625 \times 10 = 15.625 \, \Omega \]

### Step 7: Final answer
Thus, the new resistance of the wire after stretching is approximately:
\[ R' \approx 15.6 \, \Omega \]