First slide

Elastic behaviour of solids

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

Moderate

A

$AE \frac{R}{r}$
B

$AE (\frac{R - r}{r})$
C

$\frac{E}{A} (\frac{R - r}{A})$
D

$\frac{Er}{AR}$

Solution

Initial length (circumference) of the ring $= 2 πr$
Final length (circumference) of the ring $= 2 πr$
Change in length $= 2 πR - 2 πr$
$Strain = \frac{Change in length}{Original length} = \frac{2 π (R - r)}{2 πr} = \frac{R - r}{r}$
$Now Young's modulus, E = \frac{F / A}{l / L} = \frac{F / A}{(R - r) / r}$
$∴ F = AE (\frac{R - r}{r})$

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