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A wooden wheel of radius R is made of two semicircular parts (see figure). The two parts are field 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 sides by  $ \Delta T $  and it just steps over the wheel. As it cools down to the surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is  $ \alpha $  and it's Young’s modulus is Y, the force that one part of the wheel applies on the other part is?

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A.  $ 2\pi SY \alpha \Delta T $ 

B.  $ Y \alpha \Delta T $ 

C.  $ \pi SY \alpha \Delta T $ 

D.  $ 2 SY \alpha \Delta T $

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Last updated date: 18th Jun 2024
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Answer
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Hint: To answer this question, use the formula for Young’s modulus. Then, substitute the equations for stress and strain in it. Strain can also be given in terms of the coefficient of linear expansion. So, substitute this expression in the formula for Young’s modulus. Rearrange this obtained expression for Young’s modulus and obtain the expression in terms of force. So, this obtained value for force will be the amount of force that one part of the wheel applies on the other part.


Formula used:

 $ Y= \dfrac {\sigma}{\epsilon} $ 

 $  \sigma= \dfrac {F}{A} $ 

 $ \epsilon= \dfrac {\Delta L}{{L}_{0}} $ 

 $ \epsilon= \alpha \Delta T $ 


Complete answer:

Given: Radius= R

 Cross-sectional area= S

Length= L

Change in the temperature = $ \Delta T $ 

Young’s modulus= Y

Coefficient of linear expansion of the metal =  $ \alpha $ 

Young’s modulus is given by,

Young’s Modulus=  $ \dfrac {Stress}{Strain} $ 

 $ Y= \dfrac {\sigma}{\epsilon} $  …(1)

Stress is given as,

 $  \sigma= \dfrac {F}{A} $  …(2)

Where, F is the applied force

A is the cross-sectional area

Strain is given as,

 $ \epsilon= \dfrac {\Delta L}{{L}_{0}} $  …(3)

Where,  $ \Delta L $  is the total elongation

$ {L}_{0} $  is the original length

Strain is also given by,

 $ \epsilon= \alpha \Delta T $  …(4)

Substituting equation. (2) and (4) in equation. (1) we get,

 $ Y = \dfrac {F}{A \alpha \Delta T} $ 

But, cross-sectional area is given by S. Therefore, the above equation becomes,

 $ Y = \dfrac {F}{S \alpha \Delta T} $ 

 $ \Rightarrow F = YS \alpha \Delta T $ 

Therefore, the force that one part of the wheel applies on the other part is  $  YS \alpha \Delta T $ .


So, the correct answer is “Option B”.


Note: There are different types of modulus of elasticity namely, Young's modulus, Bulk modulus and Shear modulus. Students should know each modulus and its properties to answer these types of questions. Young's modulus and Shear modulus are applicable for solids while the Bulk modulus is only applicable to liquids. Due to Young's modulus and Shear modulus the shape of the material changes. While the Bulk modulus causes a change in the volume of the object.