Answer:
The maximum length of specimen is 0.475 m
Explanation:
Given :
Maximum elongation [tex]\Delta l = 0.25 \times 10^{-3}[/tex] m
Tensile force [tex]F = 8900[/tex] N
Diameter [tex]d =10.2 \times 10^{-3}[/tex] m
Elastic modulus [tex]E = 207 \times 10^{9}[/tex] Pa
So area of cylindrical specimen is given by,
[tex]A = \pi \frac{d^{2} }{4}[/tex]
From the formula of elongation,
[tex]\frac{F}{A} = \frac{\Delta l E}{l}[/tex]
Where [tex]l =[/tex] maximum length of specimen,
[tex]l = \frac{\Delta l EA}{F}[/tex]
[tex]l = \frac{\Delta l \pi E d^{2} }{4F}[/tex]
Put the value and find maximum length,
[tex]l = \frac{0.25 \times 10^{-3} \times 207 \times 10^{9} \times \pi (10.2 \times 10^{-3}) ^{2} ) }{4 \times 8900}[/tex]
[tex]l = 0.475[/tex] m
