Question

A wire of length L is hanging from a fixed support. The length changes to L₁ and L₂ when masses 1 kg and 2 kg are suspended, respectively, from its free end. Then the value of L is equal to

• A. \(\sqrt{L_1L_2}\)
• B. \(\frac{L_1+L_2}{2}\)
• C. 2L₁ – L₂
• D. 3L₁ – 2L₂

Answer 1

The Correct Option is C

To find the original length \(L\) of the wire, we need to analyze how the wire stretches under different weights.

Let's consider the problem given:

• The wire stretches to length \(L_1\) when a 1 kg mass is suspended.
• The wire stretches to length \(L_2\) when a 2 kg mass is suspended.

The problem involves the elasticity of the wire. Here, assuming the wire obeys Hooke's Law, the increase in length due to load is proportional to the load applied.

The elongation (change in length) of the wire can be expressed by Hooke’s Law as follows:

\Delta L = \frac{F}{k}

where \(F\) is the force (weight of the mass) and \(k\) is the spring constant of the wire.

Thus for masses 1 kg and 2 kg:

1. When a 1 kg mass is attached, the tension is \(T = mg = 1 \times 9.8 = 9.8\, \text{N}\), and the new length is \(L_1\).
2. When a 2 kg mass is attached, the tension is \(T = mg = 2 \times 9.8 = 19.6\, \text{N}\), and the new length is \(L_2\).

The increase in length of the wire is directly proportional to the force applied. Therefore, the increase in length from the original length can be written as:

• Increase in length when 1 kg is used: \(L_1 - L = \Delta L_1 = \frac{9.8}{k}\)
• Increase in length when 2 kg is used: \(L_2 - L = \Delta L_2 = \frac{19.6}{k}\)

By subtracting these two equations, we get the relationship:

L_2 - L_1 = \frac{19.6 - 9.8}{k} \rightarrow L_2 - L_1 = \frac{9.8}{k}

This implies that the increase in length for an additional 1 kg load is also \( \Delta L = \frac{9.8}{k}\).

Thus, combining these results, we calculate:

L = 2L_1 - L_2

Therefore, the correct answer is 2L₁ – L₂.