Question

A metal ring of initial radius $r$ and cross-sectional area $A$ is fitted onto a wooden disc of radius $R>r$. if Young's modulus of metal is $Y$ then tension in the ring is:

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

Answer 1

The correct option is B $\frac{AY(R-r)}{r}$

$\begin{array}{c}\text{Given, metal ring of radius}=r\text{and}\\ \text{cross sectional area}=A\\ l_{\text{initial}}=2\pi r\\ l_{\text{binal}}=2\pi R\end{array}$

$\begin{array}{c}\Delta l=\text{change in length}=2\pi (R-r)\\ \text{Thermal strain}=\frac{\Delta l}{l_i}=\frac{R-\gamma }{r}\\ \therefore F=yA\left(\frac{\Delta l}{l_i}\right)=yA\left(\frac{R-r}{r}\right)\end{array}$

$\therefore \text{Tension in the ring is}yA\left(\frac{R-\gamma }{r}\right)$