First slide

Elastic Modulie

Question

A 5 m long aluminium wire (Y = $7 \times 10^{10} N / m^{2}$ ) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire ( $Y = 12 \times 10^{10} N / m^{2}$ ) of the same length under the same weight, the diameter should now be, in mm.

Moderate

A

1.75
B

1.5
C

2.3
D

5.0

Solution

$l = \frac{FL}{{πr}^{2} Y} \Rightarrow r^{2} \propto \frac{1}{Y} (F, L and l are constant)$
$\frac{r_{2}}{r_{1}} = {(\frac{Y_{1}}{Y_{2}})}^{1 / 2} = {(\frac{7 \times 10^{10}}{12 \times 10^{10}})}^{1 / 2}$
$\Rightarrow r_{2} = 1 .5 \times {(\frac{7}{12})}^{1 / 2} = 1 .145 mm$
$∴ diameter = 2 .29 mm$

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