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Question

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 2πR. To fit the ring on the wheel, it is heated so that its temperature rises by ΔT 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 α and its Young's modulus is U, the force that one part of the wheel applies on the other part is?

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A
SYαΔT
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B
πSYαΔT
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C
2SYαΔT
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D
2πSYαΔT
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Solution

The correct option is A SYαΔT
Radius =R
Cross-sectional area =S
Length =L
Temperature change =T
Y is Young's Modulus
Coefficient of Linear expansion =α
Now, for the force that one part of the wheel applies on the other part is,
Y=StressStrain=FlAl(1)strain=ll=αT
Putting values in equation (1)
Y=FSαTF=SYαT
From here, force is given.

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