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

We begin by analyzing the problem using Hooke's Law, which states that the extension of a wire is proportional to the force applied, i.e., \( F = k \cdot \Delta l \), where \( F \) is the force in newtons, \( \Delta l \) is the extension, and \( k \) is the stiffness constant.
Given:
\( l_1 = 11 \times l_0 \) when \( F_1 = 100 \, \text{N} \).
\( l_1 = l_0 + \Delta l_1 \), so \( \Delta l_1 = l_1 - l_0 = 10 \, l_0 \).
For \( F_2 = 120 \, \text{N} \), let the length be \( l_2 \), then \( l_2 = l_0 + \Delta l_2 \).
The relationships of forces are:
\( 100 = k \cdot 10 \, l_0 \).
\( k = \frac{100}{10 \, l_0} = \frac{10}{l_0} \).
\( 120 = k \cdot (l_2 - l_0) \).
Substituting \( k \), we get:
\( 120 = \frac{10}{l_0} \cdot (l_2 - l_0) \).
Simplifying gives:
\( 120 = 10 \cdot \frac{l_2 - l_0}{l_0} \).
\( 12 \cdot l_0 = l_2 - l_0 \).
\( l_2 = 13 \cdot l_0 \).
From the given, \( l_1 = 11 \cdot l_0 = \frac{1}{x} \cdot l_1 \), thus:
\( \frac{11 \cdot l_0}{l_1} = \frac{1}{x} \).
Solving gives \( x = 2 \).
Checking within the range (2,2), the value 2 lies precisely there.
Therefore, \( x = 2 \) as verified against the range.