Hot water kept in a beaker placed in a room cools from `70^(@)C` to `60^(@)C` in `4` minutes. The time taken by it to cool from `69^(@)C` to `59^(@)C` from

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