Question

A wire of resistance \( X \, \Omega \) is gradually stretched till its length becomes twice its original length. If its new resistance becomes 40 \( \Omega \), find the value of \( X \).

Hint:

When a wire is stretched, its length increases, and its area decreases. The resistance increases by a factor of the square of the length change.

Answer 1

The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross
-sectional area of the wire. When the wire is stretched, its length doubles, i.e., the new length becomes \( 2L \). As the volume of the wire remains constant, the cross
-sectional area \( A \) will change. Since volume \( V = A L \), and the volume remains the same, we have: \[ A_2 L_2 = A_1 L_1 \] \[ A_2 = \frac{A_1}{2} \] This means that the new area is half the original area. Now, the new resistance \( R_2 \) is given by: \[ R_2 = \rho \frac{2L}{A_2} = \rho \frac{2L}{\frac{A_1}{2}} = 4 \rho \frac{L}{A_1} = 4R_1 \] Thus, the new resistance is four times the original resistance. Given that the new resistance is 40 \( \Omega \), we can write: \[ 4R_1 = 40 \] \[ R_1 = 10 \, \Omega \] Therefore, the original resistance \( X \) is 10 \( \Omega \).