A wire has a resistance of `10Omega`. It is stretched by `1//10` of its original length. Then its resistance will be

Options

A. `9Omega`
B. `10Omega`
C. `11Omega`
D. `12.1Omega`

Answer

To solve the problem, we need to determine the new resistance of a wire after it has been stretched. Here’s a step-by-step solution:

### Step 1: 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:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material (constant),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Analyze the effect of stretching the wire
When the wire is stretched, its volume remains constant. Therefore, if the length increases, the cross-sectional area must decrease to keep the volume constant.

### Step 3: Set up the relationship between the original and new dimensions
Let:
- \( L_1 \) = original length of the wire,
- \( A_1 \) = original cross-sectional area,
- \( R_1 = 10 \, \Omega \) (given),
- \( L_2 \) = new length after stretching,
- \( A_2 \) = new cross-sectional area.

The volume of the wire before and after stretching is given by:
\[ V = A_1 L_1 = A_2 L_2 \]
From this, we can express \( A_2 \) in terms of \( A_1 \) and the lengths:
\[ A_2 = \frac{A_1 L_1}{L_2} \]

### Step 4: Determine the new length after stretching
The wire is stretched by \( \frac{1}{10} \) of its original length. If we assume \( L_1 = 10 \) units (for simplicity), then:
\[ \text{Increase in length} = \frac{1}{10} \times L_1 = 1 \text{ unit} \]
Thus, the new length \( L_2 \) is:
\[ L_2 = L_1 + \text{Increase} = 10 + 1 = 11 \text{ units} \]

### Step 5: Set up the relationship for resistance
Using the relationship of resistance based on length and area:
\[ \frac{R_1}{R_2} = \frac{L_1^2}{L_2^2} \]
Substituting the known values:
\[ \frac{10}{R_2} = \frac{10^2}{11^2} \]
This simplifies to:
\[ \frac{10}{R_2} = \frac{100}{121} \]

### Step 6: Solve for the new resistance \( R_2 \)
Cross-multiplying gives:
\[ 10 \cdot 121 = 100 \cdot R_2 \]
\[ 1210 = 100 R_2 \]
\[ R_2 = \frac{1210}{100} = 12.1 \, \Omega \]

### Final Answer
Thus, the new resistance of the wire after stretching is:
\[ \boxed{12.1 \, \Omega} \]