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Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousMedium2 minQB

PhysicsMediumsingle choice

A thin steel ring of inner diameter $40 cm$ and cross-sectional area $1 mm^{2}$ is heated until it easily slides on a rigid cylinder of diameter $40.05 cm$ . [for steel $Y = 200 GPa$ ] When the ring cools down, the tension in the ring will be :

Options:

Answer:

Solution:

Strain in the ring $= \frac{40.05 - 40}{40} = 1.25 \times 1 0^{- 3}$

Tension $T = Y A \times strain$

$T = 200 \times 1 0^{9} \times 1 0^{- 6} \times 1.25 \times 1 0^{- 3}$

$T = 250 N$

Stream:JEESubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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