An object is cooled from `75^(@)C` to `65^(@)C` in 2 min in a room at `30^(@)C`. The time taken to cool the same object from `55^(@)C` to `45^(@)C` in the same room is

Options

A. 4
B. 5
C. 6
D. 7

Answer

A

Explanation

According to Newton's law of cooling,
`-(d theta)/(dt)prop(theta-theta_(0))rArr-(d theta)/(dt)=K(theta-theta_(0))`
Where , K is constant of proportionality .
`"Now "(75-65)/(2)=K((75+65)/(2)-30)=Kxx40`
`rArr" " K(1)/(8)`
In case of second identical object `(55-45)/(t)=K((55+45)/(2)-30)`
`rArr" " (10)/(t)=(1)/(8)xx20rArr(80)/(20)=4"min"`