Question

A 13 cm long tendon was found to stretch 3.7 mm by a force of 12.1 N . The tendon was approximately round with an average diameter of 8.8 mm . Calculate the Young's modulus of this tendon.

Answer 1

Answer:

young's modulus$=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2$

Explanation:

We have given length of tendon L= 13 cm =0.13 m

Change in length $\Delta L=3.7mm=3.7\times 10^{-3}m$

Force = 12.1 N

Average diameter d =8.8 mm

So $r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m$

Area $A=\pi\times (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2$

Now stress $=\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2$

Strain $=\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}$

Now young's modulus$=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2$