In a room where the temperature is `30^(@)C`, a body cools from `61^(@)C` to `59^(@)C` in 4 minutes. The time (in min.) taken by the body to cool from `51^(@)C` to `49^(@)C` will be

Options

A. 4min
B. 5 min
C. 6 min
D. 8 min

Answer

B

Explanation

We know that according to Newton's law of cooling
`log((theta_(1)-theta_(2))/(theta_(2)-theta_(1))) = -Kt trArrlog((61-59)/(59-(61+55)/2)) = -Kxx4`.....(i)
`rArr log((51-49)/(49-(51+49)/2))=-K xx t`.........(ii)
Solving Eqs. (i) and (ii), we get
t=5 min