First slide

Elastic behaviour of solids

Question

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1:4, the ratio of their diameters is

Moderate

A

$\sqrt{2} : 1$
B

$1 : \sqrt{2}$
C

2:1
D

$1 : 2$

Solution

$E n e r g y = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$
$\frac{E n e r g y}{v o l u m e} = \frac{1}{2} \times s t r e s s \times s t r a i n$
$E n e g y D e n s i t y = E = \frac{1}{2} \times \frac{{(s t r e s s)}^{2}}{Y} (Y = \frac{s t r e s s}{s t r a i n})$
$\Rightarrow E = \frac{1}{2} \times \frac{{(\frac{F}{A})}^{2}}{Y} = \frac{F^{2}}{2 A^{2} Y}$
$∴ \frac{E_{1}}{E_{2}} = \frac{d_{2}^{4}}{d_{1}^{4}} (∵ A = \frac{π d^{2}}{4})$
$\Rightarrow \frac{1}{4} = {(\frac{d_{2}}{d_{1}})}^{4}$
$\Rightarrow {(\frac{d_{1}}{d_{2}})}^{4} = \frac{4}{1}$
$\Rightarrow \frac{d_{1}}{d_{2}} = \frac{\sqrt{2}}{1}$

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