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AIEEE 2012 Solved Paper
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© examsnet.com
Question : 40
Total: 89
A liquid in a beaker has temperature
θ
(
t
)
at time
t
and
θ
0
is temperature of surroundings, then according to Newton's law of cooling the correct graph between
log
e
(
θ
−
θ
0
)
and
t
is:
Validate
Solution:
Newton's law of cooling
d
θ
d
t
=
−
k
(
θ
−
θ
0
)
⇒
d
θ
(
θ
−
θ
0
)
=
−
k
dt
Intergrating
⇒
log
(
θ
−
θ
0
)
=
−
k
t
+
c
Which represents an equation of straight line.
Thus the option
(
a
)
is correct.
© examsnet.com
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