A wire with 15ohm resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will be

Options

A. `15.18Omega`
B. `81.15Omega`
C. `51.18Omega`
D. `18.15Omega`

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Identify the initial conditions
Given:
- Initial resistance \( R_1 = 15 \, \Omega \)
- The wire is stretched by one-tenth of its original length.

Let the original length of the wire be \( L_1 \). After stretching, the new length \( L_2 \) becomes:
\[ L_2 = L_1 + \frac{1}{10}L_1 = \frac{11}{10}L_1 \]

### Step 3: Use the volume conservation condition
Since the volume of the wire is kept constant, we have:
\[ V_1 = V_2 \implies A_1 L_1 = A_2 L_2 \]
Where \( A_1 \) and \( A_2 \) are the cross-sectional areas before and after stretching, respectively.

### Step 4: Relate the areas using the lengths
Substituting \( L_2 \):
\[ A_1 L_1 = A_2 \left(\frac{11}{10}L_1\right) \]
This simplifies to:
\[ A_2 = \frac{A_1}{\frac{11}{10}} = \frac{10}{11}A_1 \]

### Step 5: Substitute into the resistance formula
Now we can find the new resistance \( R_2 \):
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{\frac{11}{10}L_1}{\frac{10}{11}A_1} \]
This simplifies to:
\[ R_2 = \rho \frac{11}{10} \cdot \frac{11}{10} \cdot \frac{L_1}{A_1} = \frac{121}{100} \cdot \rho \frac{L_1}{A_1} = \frac{121}{100} R_1 \]

### Step 6: Calculate the new resistance
Now substituting \( R_1 = 15 \, \Omega \):
\[ R_2 = \frac{121}{100} \cdot 15 = 18.15 \, \Omega \]

### Final Answer
The new resistance after stretching the wire is:
\[ R_2 = 18.15 \, \Omega \]

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