A metallic ring of radius $r$ and cross sectional area $A$ is fitted into a wooden circular disc of radius…

A metallic ring of radius $r$ and cross sectional area $A$ is fitted into a wooden circular disc of radius $R(R>r)$. If the Young's modulus of the material of the ring is $Y$, the force with which the metal ring expands is
  1. $\frac{A Y R}{r}$
  2. $\frac{A Y(R-r)}{r}$
  3. $\frac{Y(R-r)}{A r}$
  4. $\frac{Y R}{A R}$

Solution

$\begin{aligned} & Y=\frac{F}{A} \cdot \frac{L}{l} \\ & \therefore \text { Force } F=\frac{A Y l}{L}=\frac{A Y[2 \pi(R-r)]}{2 \pi r} \\ & \Rightarrow \quad F=\frac{A Y(R-r)}{r}\end{aligned}$

Asked in: AP EAMCET 2004

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