A steel ring of radius r and cross-section area ‘A’ is fitted on to a wooden disc of radius $R_0(R > r)$ . If Young's modulus be E, then the force with which the steel ring is expanded is
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Initial length (circumference) of the ring = 2 π π r
Final length (circumference) of the ring = 2 π π R
Change in length = 2 π π R – 2 π π r.
$\text{strain} = \frac{\text{change in length}}{\text{original length}}$ $\frac{2 \pi \left( R - r \right)}{2 \pi r}$ $= \frac{R - r}{r}$
Now Young's modulus $E = \frac{F/A}{I/L} = \frac{F/A}{(R-r)/r}$
∴ ∴ $F = AE \left(\frac{R - r}{r}\right)$
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