A wire of a certain material is stretched slowly by ten perent. Its new resistance resistance become respectively

Options

A. 1.2 times, 1.1 times
B. 1.21 times, same
C. both remain the same
D. 1.1 time, 1.1 times

Explanation

To solve the problem of how the resistance of a wire changes when it is stretched by 10%, 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:
\[ R = \rho \frac{L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the specific resistance (or resistivity) of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 2: Determine the new length after stretching
When the wire is stretched by 10%, the new length \( L' \) can be calculated as:
\[ L' = L + 0.1L = 1.1L \]

### Step 3: Apply the volume conservation principle
Since the volume of the wire remains constant during stretching, we can express this as:
\[ V = A \cdot L = A' \cdot L' \]
where \( A' \) is the new cross-sectional area. Thus, we have:
\[ A \cdot L = A' \cdot (1.1L) \]
Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ A = 1.1A' \]
This implies:
\[ A' = \frac{A}{1.1} \]

### Step 4: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length and new area:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1^2L}{A} = 1.21 \left(\rho \frac{L}{A}\right) = 1.21 R \]
where \( R \) is the original resistance.

### Step 5: Conclusion
Thus, the new resistance \( R' \) is:
\[ R' = 1.21 R \]
This means that the resistance increases by a factor of 1.21 when the wire is stretched by 10%.

### Summary of Results:
- The specific resistance \( \rho \) remains constant.
- The new resistance becomes \( 1.21 \) times the old resistance.