Question

A horizontal aluminium rod of diameter $8cm$ projects $10cm$ 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 _______. $(g=10m/s^2)$

Answer 1

Answer: 12.5

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