Question

The length of a wire becomes \( l_1 \) and \( l_2 \) when 100 N and 120 N tensions are applied respectively. If \( l_1 = 11 \, \text{l}_0 \), the natural length of the wire will be \( \frac{1}{x} \, l_1 \). Here the value of \( x \) is _____

Hint:

To determine the natural length of a wire under tension, use Hooke’s Law and consider the elongation for different applied forces.

Answer 1

Correct Answer: 2

The force in a wire is given by Hooke's Law: \[ F = kx \] Where:
- \( F \) is the applied force,
- \( k \) is the spring constant, and
- \( x \) is the elongation in the wire.
From the problem statement, we are given that: \[ F = k \Delta x \] Now, for tensions \( F_1 = 100 \, \text{N} \) and \( F_2 = 120 \, \text{N} \), the corresponding lengths of the wire will be: For \( F_1 \): \[ 100 = k (l_1 - l_0) \quad \text{(1)} \] For \( F_2 \): \[ 120 = k (l_2 - l_0) \quad \text{(2)} \] Now, let the natural length of the wire be \( l_0 \). From the given information, \( l_1 = 11l_0 \).
From equation (1), we can write: \[ 100 = k(11l_0 - l_0) \Rightarrow k(10l_0) = 100 \] Thus, \[ k = \frac{10}{l_0} \] Substituting into equation (2): \[ 120 = \frac{10}{l_0} (l_2 - l_0) \] \[ l_2 - l_0 = \frac{120 l_0}{10} = 12 l_0 \] Thus, \[ l_2 = 13 l_0 \] The ratio of \( l_2 \) to \( l_1 \) is: \[ \frac{l_2}{l_1} = \frac{13 l_0}{11 l_0} = \frac{13}{11} \] Therefore, \( x = 2 \).