Question

a wire of cross-sectional area of 0.00006m^2 and lenght 50cm stretches by 0.2mm under a load of 3000N. calculate young's modulus for the wire.

Answer 1

A wire of cross-sectional area of 0.00006m^2 and length 50cm stretches by 0.2mm under a load of 3000N. Calculate young's modulus for the wire.

Solution.

$$S = 0.00006m^2 = 6 \cdot 10^{-5}m^2, l_0 = 50cm = 0.50m, \Delta l = 0.2mm = 0.2 \cdot 10^{-3}m, F = 3000N$$ $$E - ?$$

Young's modulus, E, can be calculated by dividing the tensile stress by the tensile strain:

$$E = \frac{\sigma}{\varepsilon}.$$ $$\sigma = \frac{F}{S} - \text{the tensile stress};$$ $$\varepsilon = \frac{\Delta l}{l_0} - \text{the tensile strain}.$$ $$E = \frac{F l_0}{S \Delta l}.$$ $$E = \frac{3000 \cdot 0.50}{6 \cdot 10^{-5} \cdot 0.2 \cdot 10^{-3}} = 1.25 \cdot 10^{11} \left(\frac{N}{m^2}\right).$$

Answer:

$$E = 1.25 \cdot 10^{11} \frac{N}{m^2}.$$