1.

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

Answer»

`0.75^(@)C`/MIN
`0.25^(@)C`/min
`1.25^(@)C`/min
`1^(@)C`/min

Solution :`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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