Answer
$1.1\times 10^6N/m^2$
Work Step by Step
We can find the shear modulus as follows:
$A=\frac{\pi d^2}{4}$
$A=\frac{3.1416(0.046)^2}{4}$
$A=1.6\times 10^{-4}m^2$
The force acting on the branch is given as
$F=mg$
$F=(22Kg)(9.8m/s^2)$
$F=215.6N$
Now $F=S(\frac{\Delta x}{L_{\circ}})A$
This can be rearranged as:
$S=\frac{F}{A}(\frac{L_{\circ}}{\Delta x})$
We plug in the known values to obtain:
$S=\frac{215.6N}{16.6\times 10^{-4}m^2}(\frac{1.1m}{0.13m})$
$S=1.1\times 10^6N/m^2$
