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 $
