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. 4 minute
B. 6 minute
C. 5 minute
D. 8 minute

Answer

B

Explanation

Rate of cooling a difFerenc in temperature
`(Delta T)/(Delta t) prop Delta theta`
`rArr``(Delta T)/(Delta t) = K Delta theta`
`rArr`` Delta T = K Delta theta . Delta T`
In first case : `Delta T= 61 - 59 = 2`
`Delta theta = 60^(@) - 30^(@) = 30^(@)`
`Delta t = 4 min`
`rArr``K=(Delta T)/(Delta theta Delta t) =(2)/(30 xx 4) =(1)/(60)`
For second case :
dT = 2
`Delta theta = 50 - 30 = 20`
`dt =(dT)/(K Delta theta)=(2)/((1)/(60)xx 20) = 6 min`