First slide

Elastic behaviour of solids

Question

The length of a wire is increased by 1 mm on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is

Moderate

A

0.25 mm
B

0.5 mm
C

2mm
D

4 mm

Solution

$\begin{array}{l} Y = \frac{F / a}{Δl / l} or Y = \frac{Fl}{aΔl} \\ or Δl = \frac{Fl}{aY} = \frac{Fl}{{πr}^{2} Y} \end{array}$
In the given problem,
$Δl \propto \frac{l}{r^{2}}$
When both I and r are doubled, $∆ l$ is halved.
${Δl}_{new} = 0 .5 mm$

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