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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 3
Total: 54
A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as
T
(
t
)
=
T
0
(
1
+
β
t
1
4
)
where β is a constant with appropriate dimensions while
T
0
is a constant with dimensions of temperature. The heat capacity of metal is
4
P
(
T
(
t
)
−
T
0
)
3
β
4
T
0
4
4
P
(
T
(
t
)
−
T
0
)
β
4
T
0
2
4
P
(
T
(
t
)
−
T
0
)
2
β
4
T
0
3
4
P
(
T
(
t
)
−
T
0
)
4
β
4
T
0
5
Validate
Solution:
P
=
m
s
d
T
d
t
=
(
H
)
d
d
t
[
T
0
(
1
+
β
t
1
4
)
]
=
H
T
0
β
4
t
−
3
4
............(1)
T
=
T
0
[
1
+
β
t
1
4
]
T
T
0
−
1
=
β
t
1
4
⇒
t
−
1
/
4
=
β
T
0
T
−
T
0
⇒
t
−
3
/
4
=
(
β
T
0
T
−
T
0
)
3
.........(2)
Substitute (2) in (1)
P
=
H
T
0
β
4
β
3
T
0
3
(
T
−
T
0
)
3
⇒
H
=
4
P
(
T
−
T
0
)
3
β
4
T
0
4
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