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A body cools at the ratio of 1.2^(@)C//m...

A body cools at the ratio of `1.2^(@)C//`min when its temperature is more than that of the surrounding by `40^(@)C`. The rate of cooling of the body when its temperature is more than that of surrounding by `25^(@)C` will be

A

`0.75^(@)C`/min

B

`0.25^(@)C`/min

C

`1.25^(@)C`/min

D

`1^(@)C`/min

Text Solution

Verified by Experts

The correct Answer is:
A
`R_(1)=K(theta_(1)-theta_(0)) and R_(2)=k(theta_(2)-theta_(0))`
`(R_(2))/(R_(1))=((theta_(2)-theta_(0)))/((theta_(1)-theta_(0)))=(25)/(40)=(5)/(8)`
`R_(2)=0.75`.
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NIKITA PUBLICATION-KINETIC THEORY OF GASES & RADIATION -MCQs (Newtons Law of Cooling)
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