Question

A steel ring of 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 force with which the steel ring is expanded, is

• A. $AY\frac{R}{r}$
• B. $AY\left(\frac{R-r}{r}\right)$
• C. $\frac{Y}{A}\frac{(R-r)}{r}$
• D. $\frac{Yr}{AR}$

Answer 1

Answer: (B)

$AY\left(\frac{R-r}{r}\right)$

Young modulus $\begin{array}{c}\left(Y\right)=\frac{FL}{Ax}\\ F=\frac{AxY}{L}\end{array}$

But length $=2\pi r$ (circumference of ring)

Change in length $\left(x\right)=2\pi R-2\pi r=2\pi r=2\pi \left(R-r\right)$

Hence force of expansions is

$F=\frac{YA\times 2\pi \left(R-r\right)}{2\pi r}=\frac{YA\left(R-r\right)}{r}$