Question

A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become.

• A. 3 times
• B. 3/2 times
• C. 4 times
• D. 2 times

Answer 1

The Correct Option is D

To address this issue, we will utilize the principle of Young's Modulus, which quantifies the relationship between stress and strain in a material. The formula for the elongation of a wire under an applied force is expressed as:

\(l = \frac{FL}{A \cdot Y}\)

where:

• \(F\) represents the applied force.
• \(L\) denotes the initial length of the wire.
• \(A\) is the cross-sectional area of the wire.
• \(Y\) is the Young's Modulus of the material.

The cross-sectional area \(A\) of a wire with radius \(r\) is calculated as:

\(A = \pi r^2\)

The initial elongation of the wire is therefore:

\(l = \frac{FL}{\pi r^2 Y}\)

Subsequently, if both the radius and the applied force are reduced by half, the new radius becomes \(\frac{r}{2}\) and the new force is \(\frac{F}{2}\).

The updated cross-sectional area \(A'\) is determined by:

\(A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}\)

The new elongation, denoted as \(l'\), is given by:

\(l' = \frac{\frac{F}{2} \cdot L}{\frac{\pi r^2}{4} \cdot Y} = \frac{FL \cdot 4}{2 \cdot \pi r^2 \cdot Y} = 2 \cdot \frac{FL}{\pi r^2 Y}\)

This indicates that the new elongation is double the original elongation. Consequently, the increase in length will be:

The correct answer is: 2 times