Question

A steel ring of radius r and cross sectional area A is fitted onto a wooden disc of radius R (R > r). If the Young’s modulus of steel is Y, then the force with which the steel ring is expanded is(a) $AY(\frac{R}{r}})$(b) $AY\frac{(R-r)}{r}}$(c) $(\frac{Y}{A})[\frac{(R-r)}{r}}]$(d) $\frac{Yr}{AR}$

Answer 1

radius of steel ring is r

cross sectional area is A

and Young's modulus is Y

steel ring is fitted onto a wooden disc of radius R ( Where R > r)

then use formula,

$Y=\frac{FL}{Ae}$

where L = 2πr and e = 2πR - 2πr

so, $Y=\frac{F\times2\pi r}{A\times2\pi(R-r)}$

or, $Y=\frac{Fr}{A(R-r)}$

hence, $F=\frac{YA(R-r)}{r}$

so, option (c) is correct choice.