1.

Hot water cools from 60^(@)C" to "50^(@)C in the first 10 min and to 42^(@)C in the next 10 min. Then the temperature of the surroundings is:

Answer»

`20^(@)C`
`30^(@)C`
`15^(@)C`
`10^(@)C`

SOLUTION :According to Newton.s law of cooling
`(theta_(2)-theta_(1))/(t) =K [(theta_(1)+theta_(2))/(2)-theta_(s)]`
where, `theta_(s)` is the temperature of the surroundings.
`(60-50)/(10) =K [(60+50)/(2)-theta_(s)]`
`1=K [55-theta_(s)]"….(i)"`
Similarly, `(50-42)/(10) =K(46-theta_(s))`
`(8)/(10)=K(46-theta_(s))"...(ii)"`
Dividing Eq. (i) by Eq. (ii), we GET
`(10)/(8) =(K(55-theta_(s)))/(K(46-theta_(s)))`
`rArr theta_(s)=10^(@)C`
So, correct choice is (d).


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