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A steel ring of radius $r$ and cross-section area ' $A$ ' is fitted on to a wooden disc of radius $R(R\gtr)$. If Young's modulus be $E$, then the force with which the steel ring is expanded is
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The correct answer is:
$A E\left(\frac{R-r}{r}\right)$
Initial length (circumference) of the ring $=2 \pi r$
Final length (circumference) of the ring $=2 \pi R$
Change in length $=2 \pi R-2 \pi r$.
strain $=\frac{\text { change in length }}{\text { original length }}=\frac{2 \pi(R-r)}{2 \pi r}=\frac{R-r}{r}$
Now Young's modulus
$E=\frac{F / A}{l / L}=\frac{F / A}{(R-r) / r}$
$\therefore F=A E\left(\frac{R-r}{r}\right)$
Final length (circumference) of the ring $=2 \pi R$
Change in length $=2 \pi R-2 \pi r$.
strain $=\frac{\text { change in length }}{\text { original length }}=\frac{2 \pi(R-r)}{2 \pi r}=\frac{R-r}{r}$
Now Young's modulus
$E=\frac{F / A}{l / L}=\frac{F / A}{(R-r) / r}$
$\therefore F=A E\left(\frac{R-r}{r}\right)$
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