Answer:
(c) For steel, 1.1 x 10⁻⁴ m; for copper, 2.1x 10⁻⁴ m
Explanation:
Given;
length of the each rod, L = 1.00 m
diameter of the each rod, d = 1.50 cm
applied tensile force for the combination, F = 4000 N.
Young modulus for steel, Ysteel = 2.0 x 10¹¹ Pa
Young modulus for copper, Ycopper = 1.1 x 10¹¹ Pa
For steel:
[tex]Y_{steel} = \frac{FL}{Ae} \\\\e = \frac{FL}{AY_{steel}}[/tex]
where;
e is the elongation
A is the area of the rod
[tex]A = \frac{\pi d^2}{4} \\[/tex]
substitute the value of A back into the equation;
[tex]e = \frac{4FL}{\pi *d^2*Y_{steel}} \\\\[/tex]
[tex]e = \frac{4(4000*1)}{\pi *(0.015^2)*(2.0*10^{11}) } \\\\e = 1.1*10^{-4} \ m[/tex]
For copper:
[tex]e = \frac{4FL}{\pi d^2 Y_{copper} } \\\\e = \frac{4(4000*1)}{\pi *(0.015^2)( 1.1*10^{11} )}\\\\e = 2.1 *10^{-4} \ m[/tex]
Thus, the correct option is 'c'
(c) For steel, 1.1 x 10⁻⁴ m; for copper, 2.1 x 10⁻⁴ m
