The resistance of a wire is `10Omega`. If it is stretched ten tmes, the resistance will be

Options

A. `1Omega`
B. `10Omega`
C. `100 Omega`
D. `1000Omega`

Explanation

To solve the problem of finding the new resistance of a wire when it is stretched ten times its original length, we can follow these steps:

### Step 1: Understand the relationship between resistance, length, and cross-sectional 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,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Determine how stretching affects length and area

When the wire is stretched to ten times its original length, the new length \( L' \) becomes:
\[ L' = 10L \]

As the wire is stretched, its volume remains constant. Therefore, if the original cross-sectional area is \( A \), the new cross-sectional area \( A' \) can be found using the volume conservation principle:
\[ V = L \cdot A = L' \cdot A' \]
Substituting \( L' \):
\[ L \cdot A = 10L \cdot A' \]
From this, we can derive:
\[ A' = \frac{A}{10} \]

### Step 3: Calculate the new resistance

Now, we can calculate the new resistance \( R' \) using the new length and area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (10L)}{\frac{A}{10}} = \frac{10 \rho L \cdot 10}{A} = 100 \cdot \frac{\rho L}{A} = 100R \]
Given that the original resistance \( R \) is \( 10 \, \Omega \):
\[ R' = 100 \times 10 \, \Omega = 1000 \, \Omega \]

### Final Answer:
The new resistance of the wire after being stretched ten times its original length is \( 1000 \, \Omega \).

---