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 E \frac{R}{r}

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A E (\frac{R - r}{r})

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\frac{E}{A} (\frac{R - r}{A})

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\frac{E r}{A R}

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Solution

The correct option is C $\frac{E}{A} (\frac{R - r}{A})$
c): Initial length (circumference) of the ring $= 2 π r$

Final length (circumference) of the ring $= 2 π R$

Change in length= $2 π R - 2 π r .$

strain= $\frac{c h a n g e i n l e n g t h}{o r i g i n a l l e n g t h} = \frac{2 π (R - r)}{2 π r} = \frac{R - r}{r}$

Now young's modulus $E = \frac{F / A}{t / L} = \frac{F / A}{(R - r) / r}$

$F = A E (\frac{R - r}{r})$

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