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 \( R = \rho \frac{L}{A} \), where \( R \) is resistance, \( \rho \) is resistivity, \( L \) is length, and \( A \) is cross-sectional area. When the wire's length doubles to \( 2L \), its volume \( V = A L \) remains constant. Thus, \( A_2 L_2 = A_1 L_1 \), leading to \( A_2 = \frac{A_1}{2} \). The new resistance \( R_2 \) is \( R_2 = \rho \frac{2L}{A_2} = \rho \frac{2L}{\frac{A_1}{2}} = 4 \rho \frac{L}{A_1} = 4R_1 \). The new resistance is four times the original resistance. Given the new resistance is 40 \( \Omega \), we have \( 4R_1 = 40 \), so \( R_1 = 10 \, \Omega \). Therefore, the original resistance \( X \) is 10 \( \Omega \).