A metallic ring of radius r and cross-section | vedclass

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(B) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{F/A}{\Delta L/L}$.Here,the original length of the ring is $L = 2 \

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

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(B) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{F/A}{\Delta L/L}$.Here,the original length of the ring is $L = 2 \pi r$.The change in length required to fit the ring onto the disc of radius $R$ is $\Delta L = 2 \pi R - 2 \pi r = 2 \pi (R - r)$.Rearranging the formula for force $F$,we get $F = \frac{Y A \Delta L}{L}$.Substituting the values,$F = \frac{Y A [2 \pi (R - r)]}{2 \pi r}$.Simplifying this,we obtain $F = \frac{A Y (R - r)}{r}$.

$A$ metallic ring of radius $r$ and cross-sectional area $A$ is fitted into a wooden circular disc of radius $R$ $(R > r)$. If the Young's modulus of the material of the ring is $Y$,the force with which the metal ring expands is

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