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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