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A steel ring of radius r and cross-sectional area A is fitted on to a wooden disc of radius R (R > r). If Young's modulus of the steel is Y, then the force with which the steel ring is expanded is:

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$\frac{A Y (R - r)}{R}$

$\frac{A Y (R - r)}{r}$

$\frac{Y (R - r)}{r}$

$\frac{A Y r}{(R - r)}$

answer is D.

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Detailed Solution

Change in length of ring = $2 π R - 2 π r$

Longitudinal strain $= \frac{2 π R - 2 π r}{2 π r} = \frac{R - r}{r}$

Young's modulus, $Y = \frac{F / A}{\frac{R - r}{r}}$

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

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