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8.Mechanical Properties of Solids
hard
A steel ring of radius $r$ and cross-section area $‘A’$ is fitted on to a wooden disc of radius $R(R > r)$. If Young's modulus be $E,$ then the force with which the steel ring is expanded is
A
$AE\frac{R}{r}$
B
$AE\left( {\frac{{R - r}}{r}} \right)$
C
$\frac{E}{A}\left( {\frac{{R - r}}{A}} \right)$
D
$\frac{{Er}}{{AR}}$
Solution
(b) Initial length (circumference) of the ring $= 2$$\pi$$r$
Final length (circumference) of the ring $= 2$$\pi$$R $
Change in length $= 2$$\pi$$R -2$$\pi$$r.$
${\rm{strain}} = \frac{{{\rm{change in length}}}}{{{\rm{original length}}}}$$ = \frac{{2\pi (R – r)}}{{2\pi r}}$$ = \frac{{R – r}}{r}$
Now Young's modulus $E = \frac{{F/A}}{{l/L}} = \frac{{F/A}}{{(R – r)/r}}$
$\Rightarrow $ $F = AE\left( {\frac{{R – r}}{r}} \right)$
Standard 11
Physics
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