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Heat and Thermodynamics question

2012 · Shift 0 · Q62
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  5. /2012 · Shift 0 · Q62

Heat and Thermodynamics question

2012 · Shift 0 · Q62

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A wooden wheel of radius RRR 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 SSS and length L.LL.LL.L is slightly less than 2πR.2\pi R.2πR. To fit the ring on the wheel, it is heated so that its temperature rises by ΔT\Delta TΔT 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\alphaα, and its Young's modulus is Y,Y,Y, the force that one part of the wheel applies on the other part is : AIEEE 2012 Physics - Heat and Thermodynamics Question 396 English
  1. A
    2πSYαΔT2\pi SY\alpha \Delta T2πSYαΔT
  2. B
    SYαΔTSY\alpha \Delta TSYαΔT
  3. C
    πSYαΔT\pi SY\alpha \Delta TπSYαΔT
  4. D
    2SYαΔT2SY\alpha \Delta T2SYαΔT
View written solutionFree

Correct answer: D

  1. Thermal expansion needed to fit the ring

The metal ring has original length LLL, which is slightly less than the circumference of the wheel, 2πR2\pi R2πR.

When heated by ΔT\Delta TΔT, its length becomes

L′=L(1+αΔT).L' = L(1+\alpha \Delta T).L′=L(1+αΔT).

It is heated just enough so that it fits over the wheel, hence

L(1+αΔT)=2πR.L(1+\alpha \Delta T)=2\pi R.L(1+αΔT)=2πR.
  1. Contraction after cooling

After placing it on the wheel, the ring cools back to the original temperature. Its natural length wants to return to LLL, but it is constrained to remain at circumference 2πR2\pi R2πR.

So the tensile strain in the ring is

strain=2πR−LL.\text{strain} = \frac{2\pi R-L}{L}.strain=L2πR−L​.

From the fitting condition above,

2πR−L=LαΔT.2\pi R-L = L\alpha \Delta T.2πR−L=LαΔT.

Therefore,

strain=αΔT.\text{strain} = \alpha \Delta T.strain=αΔT.
  1. Stress in the ring

By Hooke's law,

stress=Y(strain)=YαΔT.\text{stress} = Y(\text{strain}) = Y\alpha \Delta T.stress=Y(strain)=YαΔT.

Hence the tension in the ring is

T=(stress)×S=SYαΔT.T = (\text{stress})\times S = SY\alpha \Delta T.T=(stress)×S=SYαΔT.
  1. Force between the two semicircular wooden parts

Consider one semicircular half of the wheel. The ring pulls on it at its two ends. At each end, the pull is tangential and has magnitude TTT.

For a semicircle, these two tangential pulls are both along the diameter and act so as to compress the two halves together. Their resultant is

F=T+T=2T.F = T + T = 2T.F=T+T=2T.

Thus,

F=2SYαΔT.F = 2SY\alpha \Delta T.F=2SYαΔT.
  1. Matching with options

So the correct option is

2SYαΔT\boxed{2SY\alpha \Delta T}2SYαΔT​

which is Option D.

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