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

Options

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

Answer

To solve the problem, we need to analyze how the resistance and specific resistance of a wire change when it is 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 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
The wire is stretched by 10%. Therefore, the new length \( L' \) can be calculated as:
\[ L' = L + 0.1L = 1.1L \]

### Step 3: Determine the new area after stretching
When the wire is stretched, its volume remains constant. The volume \( V \) of the wire is given by:
\[ V = L \times A \]
After stretching, the new volume \( V' \) is:
\[ V' = L' \times A' = 1.1L \times A' \]
Setting the initial and final volumes equal gives:
\[ L \times A = 1.1L \times A' \]
From this, we can solve for the new area \( A' \):
\[ A' = \frac{A}{1.1} = \frac{10}{11}A \]

### Step 4: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length and new area:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (1.1L)}{\frac{10}{11}A} \]
This simplifies to:
\[ R' = \frac{\rho L}{A} \times \frac{1.1 \times 11}{10} = R \times \frac{12.1}{10} = 1.21R \]

### Step 5: Conclusion
Thus, the new resistance \( R' \) is \( 1.21 \) times the initial resistance \( R \), and the specific resistance \( \rho \) remains unchanged.

### Final Answer:
- New Resistance: \( R' = 1.21R \)
- Specific Resistance: \( \rho' = \rho \)

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