Answer:
(a) Aluminium= [tex]6.3492*10^{7} N/m^{2}[/tex] , steel=[tex]6.366*10^{7} N/m^{2}[/tex]
(b) Aluminium=[tex]0.0907*10^{-2}[/tex] , steel=[tex]0.0318*10^{-2}[/tex]
(c) Aluminium= 0.2721 cm, Steel= 0.009549cm
Explanation:
(a)
For aluminium bar, stress, [tex]\sigma[/tex] is given by
[tex]\sigma= \frac {F}{A}[/tex] where A is cross-sectional area of bar and F is force applied
Therefore, [tex]\sigma= \frac {20*10^{3}}{1.5*10^{-2}*2.1*10^{-2}}=6.3492*10^{7} N/m^{2}[/tex]
For steel rod,
[tex]\sigma= \frac {F}{A}= \frac {20*10^{3}}{ \pi R^{2}}=\frac {20*10^{3}}{ \pi 0.01^{2}}=6.366*10^{7} N/m^{2}[/tex]
(b)
Strain, [tex]\epsilon[/tex] in aluminium is given by
[tex]\epsilon = \frac { \sigma}{E} [/tex]where E is Young’s Modulus
[tex]\epsilon= \frac {6.39*10^{7}}{70*10^{9}}=0.0907*10^{-2}[/tex]
For steel rod
[tex]\epsilon = \frac { \sigma}{E}[/tex]
[tex]\epsilon= \frac {6.37*10^{7}}{200*10^{9}}=0.0318*10^{-2}[/tex]
(c)
Strain, [tex]\epsilon[/tex] is given by
[tex]\epsilon = \frac {Elongation}{original length}[/tex] hence the change in length is product of original length and [tex]\epsilon[/tex]
For aluminium rod,
[tex]Elongation= 30*0.0907*10^{-2}=0.2721 cm[/tex]
For steel rod
[tex]Elongation= 30*0.0318*10^{-2}= 0.009549 cm[/tex]
