The resistance of a wire is `10 Omega`. Its length is increased by `10%` by stretching. The new resistance will now be

Options

A. `12Omega`
B. `1.2Omega`
C. `13Omega`
D. `11Omega`

Answer

To find the new resistance of a wire after it has been stretched, we can follow these steps:

### Step 1: Understand the initial conditions
The initial resistance \( R_1 \) of the wire is given as \( 10 \, \Omega \).

### Step 2: Determine the change in length
The length of the wire is increased by \( 10\% \). If the original length of the wire is \( l \), then the new length \( l' \) after stretching is:
\[ l' = l + 0.1l = 1.1l \]

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

### Step 4: Analyze the effect of stretching on area
When the wire is stretched, its volume remains constant. Therefore, if the length increases, the cross-sectional area must decrease. The volume \( V \) of the wire is given by:
\[ V = l \cdot A \]
After stretching, the volume remains the same:
\[ l \cdot A = l' \cdot A' \]
Substituting \( l' = 1.1l \):
\[ l \cdot A = 1.1l \cdot A' \]
From this, we can solve for the new area \( A' \):
\[ A' = \frac{A}{1.1} \]

### Step 5: Calculate the new resistance
Now we can express the new resistance \( R_2 \) using the new length and area:
\[ R_2 = \rho \frac{l'}{A'} = \rho \frac{1.1l}{\frac{A}{1.1}} = \rho \frac{1.1^2 l}{A} = 1.1^2 \cdot \rho \frac{l}{A} = 1.1^2 R_1 \]
Substituting \( R_1 = 10 \, \Omega \):
\[ R_2 = 1.1^2 \cdot 10 \, \Omega = 1.21 \cdot 10 \, \Omega = 12.1 \, \Omega \]

### Step 6: Final result
Thus, the new resistance of the wire after stretching is approximately:
\[ R_2 \approx 12 \, \Omega \]