A steel ring of radius 'r' is to be fitted over a wooden disc of radius 'R' $(\mathrm{R}>\mathrm{r})$. The…

A steel ring of radius 'r' is to be fitted over a wooden disc of radius 'R' $(\mathrm{R}>\mathrm{r})$. The force required to expand the ring so that it fits over the disc is $[\mathrm{Y}=$ Young's modulus of steel, $\mathrm{A}=$ area of cross section of wire $]$
  1. $\mathrm{YA}\left(\frac{\mathrm{R}-\mathrm{r}}{\mathrm{r}}\right)$
  2. $\mathrm{YA}\left(\frac{\mathrm{r}}{\mathrm{R}-\mathrm{r}}\right)$
  3. $\mathrm{YA} \frac{\mathrm{r}}{\mathrm{R}}$
  4. $\left(\frac{\mathrm{YAR}}{\mathrm{r}}\right)$

Solution

$(\mathrm{Y})=\frac{\mathrm{F} \mathrm{L}}{\mathrm{Ax}}$ Young modulus $\mathrm{F}=\frac{\mathrm{AxY}}{\mathrm{L}}$ But length $=2 \pi r$ (circumference of ring) Change in length $(\mathrm{x})=2 \pi \mathrm{R}-2 \pi \mathrm{r}=2 \pi \mathrm{r}=2 \pi(\mathrm{R}-\mathrm{r})$ Hence force of expansions is $\mathrm{F}=\frac{\mathrm{YA} \times 2 \pi(\mathrm{R}-\mathrm{r})}{2 \pi \mathrm{r}}=\frac{\mathrm{YA}(\mathrm{R}-\mathrm{r})}{\mathrm{r}}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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