Question

A horizontal aluminium rod of diameter $8 cm$ projects $10 cm$ from a wall. An object of mass $600 \pi kg$ is suspended from the end of the rod. The shear modulus of aluminium is $3 \times 10^{10} N / m ^{2}$. The vertical deflection of the end of the rod in $\mu m$ is _______. $(g = 10\,m/s^2)$

Answer 1

$ \eta =\frac{ FL }{ Al } $
$ \therefore l =\frac{ FL }{ A \eta}=\frac{ mgL }{\pi\left(\frac{ d }{2}\right)^{2} \eta} $
$=\frac{600 \pi \times 10 \times 10 \times 10^{-2}}{\pi \times\left(4 \times 10^{-2}\right)^{2} \times 3 \times 10^{10}}=\frac{25}{2} \times 10^{-6}$
$=12.5 \times 10^{-6} m$
$\therefore l = 12.5\,\mu m$