Properties Of Solids And Liquids Question 421

Question: 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,$ \alpha =1{{0}^{-5}}/{}^\circ C, Y= 200 GPa $ ] When the ring cools down, the tension in the ring will be:

Options:

A) 1000 N

B) 500 N

C) 250 N

D) 100 N

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $ \frac{\Delta \ell }{\ell }=\frac{\pi \times 0.5}{\pi \times 40}=\frac{1}{800} $

$ T=Y\frac{\Delta \ell }{\ell }\times A $

$ =200\times 10^{9}\times \frac{1}{800}\times 1\times {{10}^{-6}}=250N $



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