Question

A thin steel ring of inner radius $r$ and cross-sectional area $A$ is fitted on to a wooden disc of radius $R(R > r)$. If Young's modulus be $Y$, then the tension in the steel ring is $A Y\left(\frac{R-r}{n r}\right)$. Find $n$.

Answer 1

Strain $=\frac{\Delta l}{l}=\left(\frac{2 \pi R-2 \pi r}{2 \pi r}\right)$
$\Rightarrow \frac{\Delta l}{l}=\left(\frac{R-r}{r}\right)$
$Y=\frac{F / A}{\frac{\Delta l}{l}}$
$ \Rightarrow Y \frac{\Delta l}{l} A=F $
$\Rightarrow F=A Y\left(\frac{R-r}{r}\right)$