A steel ring of radius $r$ and cross-section area $'A'$ is fitted on to a wooden disc of radius $R(R r)$. If Young’s modulus be $E$, then the force with which the steel ring is expanded is

Question

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

Dr. Abubacker · Accepted Answer

(b) Initial length (circumference) of the ring $= 2\pi r$
Final length (circumference) of the ring $= 2\pi R$
Change in length $= 2\pi R - 2\pi r$
$strain = \frac{change\ in\ length}{original\ length} = \frac{2\pi(R - r)}{2\pi r} = \frac{R - r}{r}$
Now Young’s modulus $E = \frac{F/A}{l/L} = \frac{F/A}{(R - r)/r}$
$	herefore F = AE\left( \frac{R - r}{r} ight)$

Q. A steel ring of radius $r$ and cross-section area $'A'$ is fitted on to a wooden disc of radius $R(R > r)$. If Young’s modulus be $E$, then the force with which the steel ring is expanded is

Explanation

Reviewed by Dr. Abubacker

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