A thin steel ring of inner diameter 40 cm and cross-sectional area 1 mm2, is heat 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 :

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Question

A thin steel ring of inner diameter 40 cm and cross-sectional area 1 mm², is heat 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 :

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$\frac{Δ l}{l} = \frac{π \times 0 .05}{π \times 40} = \frac{1}{800}$

$T = Y \frac{Δ l}{l} \times A = 200 \times 10^{9} \times \frac{1}{800} \times 1 \times 10^{- 6} = 250 N$ .

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