A wire with 15 ohmresistance 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 need to determine the new resistance of a wire after it has been stretched while keeping its volume constant. Let's break this down step by step.

### Step 1: Understand the initial conditions
Let the initial resistance of the wire be \( R_1 = 15 \, \Omega \). Let the original length of the wire be \( l \) and the original radius be \( r \).

### Step 2: Determine the new length after stretching
The wire is stretched by one-tenth of its original length. Therefore, the new length \( l_2 \) can be calculated as:
\[ l_2 = l + \frac{1}{10}l = \frac{11}{10}l \]

### Step 3: Use the constant volume condition
The volume of the wire before stretching is given by:
\[ V = \pi r^2 l \]
After stretching, the volume remains constant, so we have:
\[ V = \pi r_2^2 l_2 \]
Setting the two volume equations equal gives:
\[ \pi r^2 l = \pi r_2^2 \left(\frac{11}{10}l\right) \]
Cancelling \( \pi \) and \( l \) (assuming \( l \neq 0 \)):
\[ r^2 = \frac{11}{10} r_2^2 \]
Rearranging gives:
\[ r_2^2 = \frac{10}{11} r^2 \]

### Step 4: Calculate the new resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho l}{A} \]
where \( A \) is the cross-sectional area. The new resistance \( R_2 \) can be expressed as:
\[ R_2 = \frac{\rho l_2}{A_2} \]
Substituting \( l_2 \) and \( A_2 \):
\[ R_2 = \frac{\rho \left(\frac{11}{10} l\right)}{\pi r_2^2} \]
Substituting \( r_2^2 \):
\[ R_2 = \frac{\rho \left(\frac{11}{10} l\right)}{\pi \left(\frac{10}{11} r^2\right)} \]
This simplifies to:
\[ R_2 = \frac{11}{10} \cdot \frac{11}{10} \cdot \frac{\rho l}{\pi r^2} = \frac{121}{100} R_1 \]
Substituting \( R_1 = 15 \, \Omega \):
\[ R_2 = \frac{121}{100} \cdot 15 = 18.15 \, \Omega \]

### Conclusion
The new resistance of the wire after stretching is \( R_2 = 18.15 \, \Omega \).