1.

A body cools from 50^(@)C to 40^(@)C in 5 minutes and from 40^(@)C to 30^(@)C in 8 minutes. Then the temperature of the surrounding is

Answer»

`25^(@)C`
`18^(@)C`
`15^(@)C`
`10^(@)C`

Solution :Here `T_(1)=0^(@)C,T_(2)+40^(@)C,t+5` min Let `T_(c)` be temp. of curroundings
`:.` average temperature `T+(T_(1)+T_(2))/(2^(@)C)`
CHANGE in temperature `dT+=0^(@)C`
`:.(dT)/(dt)+_k(T-T_(c))RARR(10)/(5)=-k(45-T_(c))`. . .(i) For `2^(ND)` case
`(10)/(8)=-k(35-T_(c))`. . . (ii)
DIVIDING (i) by (ii)
`(8)/(5)=(45-T_(c))/(35-T_(c))rArr8xx35-8T_(c)=5xx45-5T_(c)`
`T_(c)~=18^(@)C`
So correct choice is (b).


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