First slide

Elastic Modulie

Question

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

Easy

A

$A E \frac{R}{r}$
B

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

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

$\frac{E r}{A R}$

Solution

Initial length ( circumference) of the ring is $2 π r$
Final length ( circumference ) of the ring is $2 π R$
Change in length = $2 π R - 2 π r$
$S t r a i n = \frac{2 π R - 2 π r}{2 π r} = \frac{R - r}{r}$
$N o w Y o u n g' s \mod u l u s E = \frac{\frac{F}{A}}{\frac{l}{L}} = \frac{F / A}{(R - r) / r} \Rightarrow F = A E (\frac{R - r}{r})$

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