A wooden wheel of radius R is made of two semicircular part (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 2π R. To fit the ring on the wheel, it is heated so that its temperature rises by Δ T and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is α, and its Young's modulus is Y, the force that one part of the wheel applies on the other part is: (a) 2π SYαΔ T (b) SYαΔ T (c) π SYαΔ T (d) 2SYαΔ T

Answer: (d). TTring half (top)net pull = 2T Ring strain after cooling: (2π R - L)/(L) = αΔ T. Tension in the strip: T = YSαΔ T. Each half of the ring pulls the two halves together with T at both cut ends, so force = 2T. F = 2SYαΔ T, answer (d).

A wooden wheel of radius R is made of two semicircular part (see… | JEE Main 2012 (AIEEE)

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Thermometry & Thermal Expansion

A wooden wheel of radius $R$ is made of two semicircular part (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 $\Delta T$ and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$ , and its Young's modulus is $Y,$ the force that one part of the wheel applies on the other part is:

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Answer: (d)

Ring strain after cooling: $\dfrac{2\pi R - L}{L} = \alpha\Delta T$ .
Tension in the strip: $T = YS\alpha\Delta T$ .
Each half of the ring pulls the two halves together with $T$ at both cut ends, so force $= 2T$ .
$F = 2SY\alpha\Delta T$ , answer (d).

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