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Question

The inner diameter of a brass ring at 273 K is 5 cm.To what temperature should it be heated for it to accommodate a ball 5)01 cm in diameter. (α=2×105/oC)

A
273 K
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B
373 K
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C
437 K
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D
173 K
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Solution

The correct option is A 373 K
In this case, d1=5 cm, d2=5.01 cm
so, d2d1=0.01 cm
dt = change in temperature = dl/(αl)) = 100
Initial temperature = 273 K
So, final temperature = 100 + 273 = 373K

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