1.

A hot body is allowed to cool. The surrounding temperature is constant at 30^(@)C. This takes time t_(1) to cool from 70^(@)C" to "68^(@)C and time t_(2) to cool from 60^(@)C" to "59.5^(@)C. Then

Answer»

`t_(2)=t_(1)`
`t_(2)=2t_(1)`
`t_(2)=(1)/(2)t_(1)`
`t_(2)=4t_(1)`

Solution :By Newton.s law of cooling
`(dT)/(dt) =ALPHA (T-T_(0))`
`therefore (2)/(t_(1)) =alpha (90-30)=60 alpha" ….(i)"`
and `(0.5)/(t_(2))=alpha (60-30)=30 alpha" …..(ii)"`
So, correct CHOICE is (b).


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