Question

A wire of unknown material with a cross-sectional area of 1 mm^2 and an unstretched length of 0.5 m is stretched through the application of known forces. From the plot of force as a function of Delta L in the figure, the Young's modulus of the material is: A. 4 times 10^10 N/m^2 B. 80 N/m^2 C. 40 N/m^2 D. 8 times 10^10 N/m^2 E. none of the above

Answer 1

A = 1 mm^2 = 10^-6 m^2

stress = Y (strain)

F / A = Y ( deltaL / L)

F = (A / L) Y deltaL

F = (10^-6 / 0.5) Y (deltaL)

F = (10^-6 x 2 Y ) (deltaL)

slope of F vs deltaL curve. = 10^-6 x 2 Y

40 / ( (0.75 - 0.25) x 10^-3) = 10^-6 x 2 Y

Y = 4 x 10^10 Pa

Ans(A)