Question

A 19

Answer 1

Answer:

Young's modulus of this tendon is $9.03\times 10^6\ N/m^2$.

Explanation:

Given that,

Length of the tendon, l = 19 cm

It is stretched by 4.5 mm, $\Delta l=4.5\ mm$

Force, F = 11.3 N

Average diameter, d = 8.2 mm

Radius, r = 4.1 mm

The formula of Young's modulus of this tendon is given by :

$Y=\dfrac{Fl}{\Delta l A}\\\\Y=\dfrac{11.3\times 0.19}{4.5\times 10^{-3}\times \pi (4.1\times 10^{-3})^2}\\\\Y=9.03\times 10^6\ N/m^2$

So, the Young's modulus of this tendon is $9.03\times 10^6\ N/m^2$. Hence, this is the required solution.