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

Options

A. `t_(2)=t_(1)`
B. `t_(2)=2t_(1)`
C. `t_(2)=(1)/(2)t_(1)`
D. `t_(2)=4t_(1)`

Answer

B

Explanation

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).