Question

Consider the following data to find the Young’s Modulus, E, of a steel wire of length l and diameter d,

given that:

E = 4 Mgl / πed 2

and:

Length of wire (l) = (3.025 ± 0.005) m

Diameter of wire (d) = (0.84 ± 0.01) mm

Mass supported by wire (M) = (5.000 ± 0.002) kg

Extension caused (e) = (1.27 ± 0.02) mm

Acceleration of free fall (g) = (9.81 ± 0.01) ms -2

Answer 1

$E=4Mgl/(\pi ed^2)=$

$=4\cdot5.000\cdot9.81\cdot3.025\cdot10^9/(3.14\cdot1.27\cdot0.84^2)=$

$=210.82\ (GPa)$

$\Delta E/E=\sqrt{(\Delta M/M)^2+(\Delta g/g)^2+(\Delta l/l)^2+(\Delta e/e)^2+(2\Delta d/d)^2}=$

$=\sqrt{( 0.002/5)^2+(0.01/9.81)^2+(0.005/3.025)^2+(0.02/1.27)^2+(2\cdot 0.01/0.84)^2}=0.0286$

$\Delta E=6.03\ (GPa)$

$E=(210.82±6.03)\ GPa$