MHT CET · Physics · Mechanical Properties of Solids
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 \(]\)
- A \(\mathrm{YA}\left(\frac{\mathrm{R}-\mathrm{r}}{\mathrm{r}}\right)\)
- B \(\mathrm{YA}\left(\frac{\mathrm{r}}{\mathrm{R}-\mathrm{r}}\right)\)
- C \(\mathrm{YA} \frac{\mathrm{r}}{\mathrm{R}}\)
- D \(\left(\frac{\mathrm{YAR}}{\mathrm{r}}\right)\)
Answer & Solution
Correct Answer
(A) \(\mathrm{YA}\left(\frac{\mathrm{R}-\mathrm{r}}{\mathrm{r}}\right)\)
Step-by-step Solution
Detailed explanation
\((\mathrm{Y})=\frac{\mathrm{F} \mathrm{L}}{\mathrm{Ax}}\)
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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