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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 $]$
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The correct answer is:
$\mathrm{YA}\left(\frac{\mathrm{R}-\mathrm{r}}{\mathrm{r}}\right)$
$(\mathrm{Y})=\frac{\mathrm{F} \mathrm{L}}{\mathrm{Ax}}$
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
$\mathrm{F}=\frac{\mathrm{YA} \times 2 \pi(\mathrm{R}-\mathrm{r})}{2 \pi \mathrm{r}}=\frac{\mathrm{YA}(\mathrm{R}-\mathrm{r})}{\mathrm{r}}$
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
$\mathrm{F}=\frac{\mathrm{YA} \times 2 \pi(\mathrm{R}-\mathrm{r})}{2 \pi \mathrm{r}}=\frac{\mathrm{YA}(\mathrm{R}-\mathrm{r})}{\mathrm{r}}$
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