Question

The resistance of a wire is \( 5 \Omega \). What will be its new resistance in ohms if stretched to 5 times its original length?

• A. \( 625 \)
• B. \( 5 \)
• C. \( 125 \)
• D. \( 25 \)

Hint:

For a wire stretched \( n \) times its original length: \[ R_{{new}} = n^2 R_{{old}} \] where \( n \) is the stretch factor.

Answer 1

The Correct Option is C

Step 1: Understanding resistance change in a stretched wire. The resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] where: - \( \rho \) is the resistivity (constant for the material), - \( L \) is the length, - \( A \) is the cross-sectional area. When a wire is stretched to \( n \) times its original length, its volume remains constant: \[ {Initial Volume} = {Final Volume} \] Since volume is \( A L \), we get: \[ A_{{new}} L_{{new}} = A_{{old}} L_{{old}} \] Given \( L_{{new}} = 5 L_{{old}} \), the new cross-sectional area becomes: \[ A_{{new}} = \frac{A_{{old}}}{5} \] Step 2: Finding new resistance. \[ R_{{new}} = \rho \frac{L_{{new}}}{A_{{new}}} = \rho \frac{5L}{A/5} \] \[ R_{{new}} = \rho \frac{5L}{A} \times 5 = 25R \] Since the original resistance is \( R = 5 \Omega \), \[ R_{{new}} = 25 \times 5 = 125 \Omega \] Final Answer: \[ \boxed{125 \Omega} \]