Question

A conducting wire is stretched by applying a deforming force, so that its diameter decreases to 40% of the original value. The percentage change in its resistance will be: 
 

• A. \(0.9\%\)
• B. \(0.12\%\)
• C. \(1.6\%\)
• D. \(0.5\%\)

Hint:

When a wire is stretched, its length increases and its cross-sectional area decreases, leading to an increase in resistance.

Answer 1

The Correct Option is C

Step 1: {Understanding the effect of stretching} 
Since the volume of the wire remains constant, we use the relation: \[ V = A l \] where \(A\) is the cross-sectional area and \(l\) is the length. 
Step 2: {Deriving the new resistance} 
We use the resistance formula: \[ R = \rho \frac{l}{A} \] Since \(A\) decreases as \(d^2\) and \(l\) increases proportionally: \[ \frac{\Delta R}{R} = -4 \frac{\Delta D}{D} \] Substituting \(\Delta D = -0.4\), \[ \frac{\Delta R}{R} = -4(-0.4) = 1.6\% \] Thus, the percentage change in resistance is \(1.6\%\).