A wooden wheel of radius R is made of two semicircular parts (see...

A wooden wheel of radius

\mathrm{R}

is made of two semicircular parts (see figure); The two parts are held together by a ring made of a metal strip of cross sectional area S and length

L

L

is slightly less than

2 \pi R

. To fit the ring on the wheel, it is heated so that its temperature rises by

\Delta \mathrm{T}

and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is

\alpha

, and its Youngs' modulus is

Y

, the force that one part of the wheel applies on the other part is : A wooden wheel of radius R is made of two semicircular parts (see figure); The two parts are held together by a ring

A wooden wheel of radius R is made of two semicircular parts (see figure); The two parts are held together by a ring

Options

A $\mathrm{2 \pi S Y \alpha \Delta T}$
B $\mathrm{S Y \alpha \Delta T}$
C $\mathrm{\pi S Y \alpha \Delta T}$
D $\mathrm{2 S Y \alpha \Delta T}$

Correct answer

D. $\mathrm{2 S Y \alpha \Delta T}$

Step-by-step solution

If temperature increases by

\Delta \mathrm{T}

, Increase in length

\mathrm{L}, \Delta

\mathrm{L}=\mathrm{L}

\alpha \Delta \mathrm{T}

\therefore \quad \frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha \Delta \mathrm{T}

Let tension developed in the ring is

\mathrm{T}

\therefore \quad \frac{\mathrm{T}}{\mathrm{S}}=\mathrm{Y} \frac{\Delta \mathrm{L}}{\mathrm{L}}=\mathrm{Y} \alpha \Delta \mathrm{T}

\therefore \quad

\mathrm{T}=\mathrm{S}

\mathrm{Y} \alpha \Delta \mathrm{T}

From FBD of one part of the wheel,

\mathrm{F}=2 \mathrm{T}

Where,

F

is the force that one part of the wheel applies on the other part.

\therefore \mathrm{F}=2 \mathrm{SY} \alpha \Delta \mathrm{T}

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