The inner diameter of a brass ring at 273K is 5cm.To what temperature should it be heated for it to accommodate a ball 5)01cm in diameter. (α=2×10−5/oC)
A
273K
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B
373K
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C
437K
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D
173K
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Solution
The correct option is A373K In this case, d1=5 cm, d2=5.01 cm so, d2−d1=0.01 cm dt = change in temperature =dl/(αl))= 100 Initial temperature = 273 K So, final temperature = 100 + 273 = 373K