Question

A wire of cross-sectional area of 6x10-5mand length 50cm stretches by 0.2mm under a load of 3000N. Calculate the Young’s modulus for the wire

Answer 1

Young's modulus, $E$, can be calculated by as:

$$E = \frac {F L 0}{A 0 \Delta L}$$

It is defined as the ratio of the uniaxial stress over the uniaxial strain in the range of stress in which Hooke's Law holds.

Where:

$E$ – is the Young's modulus (modulus of elasticity)

$F$ – ie the force exetered on an object

$A0$ – is the original cross-sectional area

$\Delta L$ – is the amount by which the length of the object changes

$L0$ – is the original length of the object

Let:

$$F = 3 0 0 0 N$$ $$A 0 = 6 * 1 0 ^ {- 5} m ^ {2}$$ $$\Delta L = 0. 2 m m = 2 * 1 0 ^ {- 4} m$$ $$L 0 = 5 0 c m = 0. 5 m$$ $$E = ?$$ $$E = \frac {F L 0}{A 0 \Delta L} = \frac {3 0 0 0 * 0 . 5}{6 * 1 0 ^ {- 5} * 2 * 1 0 ^ {- 4}} = 1 2 5 \mathrm {G P a} (1 2 5 * 1 0 ^ {9} \mathrm {P a} - \text {p a s c a l o r N / m} ^ {2}).$$

Answer:

$$E = 1 2 5 G P a$$