Question

A wire of a certain material is stretched slowly by 10%. Its new resistance and specific resistance becomes respectively

• A. 1.21 times, same
• B. 1.1 times, 1.1 times
• C. both remains the same
• D. 1.2 times, 1.1 times

Answer 1

The Correct Option is A

Let the original length be \( L \) and area be \( A \). 
When the wire is stretched by 10%, new length becomes:  
\[ L' = 1.1L \] 
Since volume remains constant: 
\[ A L = A' L' \Rightarrow A' = \frac{A}{1.1} \] 
So, new resistance is: 
\[ R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{A / 1.1} = \rho \frac{1.1^2 L}{A} = 1.21 R \] 
Resistance becomes 1.21 times. 

Specific resistance (resistivity) \( \rho \) depends only on the material, not on the dimensions, so it remains same. 

Correct answer: 1.21 times, same

Answer 2

When a wire is stretched, its resistance and specific resistance may change. Let's analyze each option to determine the correct answer:

If the new resistance becomes 1.21 times and the specific resistance remains the same, this implies that the wire's length increased while its cross-sectional area remained constant. 

Therefore, the correct answer is: (A) the new resistance becomes 1.21 times and the specific resistance remains the same.