A wire of `1Omega` has a length of 1m. It is stretched till its length increases by 25%. The percentage change in resistance to the neartest integer is:-

Options

A. `56%`
B. `25%`
C. `12.5%`
D. `76%`

Answer

To solve the problem, we need to determine the percentage change in resistance when a wire is stretched. Here are the steps to find the solution:

### Step 1: Understand the initial parameters
- The initial resistance \( R_1 = 1 \, \Omega \)
- The initial length \( L_1 = 1 \, m \)
- The initial radius \( r_1 \) is unknown but will be used in calculations.

### Step 2: Calculate the new length after stretching
- The wire is stretched by 25%, so the new length \( L_2 \) is:
\[ L_2 = L_1 + 0.25 \times L_1 = 1 + 0.25 \times 1 = 1.25 \, m \]

### Step 3: Use the formula for resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

### Step 4: Express the area in terms of radius
The cross-sectional area \( A \) of the wire can be expressed as:
\[ A = \pi r^2 \]

### Step 5: Determine the new area after stretching
When the wire is stretched, its volume remains constant. Therefore:
\[ L_1 A_1 = L_2 A_2 \]
Substituting the known values:
\[ 1 \cdot A_1 = 1.25 \cdot A_2 \implies A_2 = \frac{A_1}{1.25} \]

### Step 6: Calculate the new resistance
Using the new length and area:
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{1.25}{\frac{A_1}{1.25}} = \rho \frac{1.25^2}{A_1} = 1.5625 \cdot R_1 \]
Since \( R_1 = 1 \, \Omega \):
\[ R_2 = 1.5625 \, \Omega \]

### Step 7: Calculate the percentage change in resistance
The percentage change in resistance is given by:
\[ \text{Percentage Change} = \frac{R_2 - R_1}{R_1} \times 100 = \frac{1.5625 - 1}{1} \times 100 = 56.25\% \]
Rounding to the nearest integer, we get:
\[ \text{Percentage Change} \approx 56\% \]

### Final Answer
The percentage change in resistance to the nearest integer is **56%**.

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