A steel ring of radius r and cross-sectional | vedclass

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(B) Young's modulus is defined as $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectiona

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$AY\,\left( \frac{R - r}{r} \right)$

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(B) Young's modulus is defined as $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the change in length.Rearranging for force,we get $F = \frac{Y \cdot A \cdot \Delta L}{L}$.The original circumference of the steel ring is $L = 2 \pi r$.The final circumference when fitted on the disc is $2 \pi R$.The change in length is $\Delta L = 2 \pi R - 2 \pi r = 2 \pi (R - r)$.Substituting these values into the force equation:$F = \frac{Y \cdot A \cdot 2 \pi (R - r)}{2 \pi r} = \frac{YA(R - r)}{r}$.

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

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