Question

Two cylindrical rods, one stainless steel and the other annealed copper, are joined end to end. Each rod is 1.00 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what is the elongation? Ysteel = 2.0 x 1011 Pa, Ycopper = 1.1 x 1011 Pa

a) For steel, 1.6 x 10m ; for copper, 8.3 x 10m
b) For steel, 2.1 x 10 m; for copper, 1.1 x 10 m
c) For steel, 1.1 x 10 m; for copper, 2.1x 10m
d) For steel, 8.3 x 10m; for copper, 1.6 x 10m

Answer 1

(c) For steel, 1.1 x 10⁻⁴ m; for copper, 2.1x 10⁻⁴ m

Step-by-step 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:

`Y_(steel) = (FL)/(Ae) \\\\e = (FL)/(AY_(steel))`

where;

e is the elongation

A is the area of the rod

`A = (\pi d^2)/(4) \\`

substitute the value of A back into the equation;

`e = (4FL)/(\pi *d^2*Y_(steel)) \\\\`

`e = (4(4000*1))/(\pi *(0.015^2)*(2.0*10^(11)) ) \\\\e = 1.1*10^(-4) \ m`

For copper:

`e = (4FL)/(\pi d^2 Y_(copper) ) \\\\e = (4(4000*1))/(\pi *(0.015^2)( 1.1*10^(11) ))\\\\e = 2.1 *10^(-4) \ m`

Thus, the correct option is 'c'

(c) For steel, 1.1 x 10⁻⁴ m; for copper, 2.1 x 10⁻⁴ m