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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 beA. `0.75^(@)C`/minB. `0.25^(@)C`/minC. `1.25^(@)C`/minD. `1^(@)C`/min |
Answer» Correct Answer - 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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