Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
A body takes 4 minutes to cool from 100° C to 70° c. If temperature of the surrounding is 20° C. The time taken by it to operatornamecool from 70° C to 50° C is
Q. A body takes 4 minutes to cool from
10
0
∘
C
to
7
0
∘
c
. If temperature of the surrounding is
2
0
∘
C
. The time taken by it to
cool
from
7
0
∘
C
to
5
0
∘
C
is
2993
209
Thermal Properties of Matter
Report Error
A
13/3
min
26%
B
13
min
7%
C
3
min
43%
D
6
min
23%
Solution:
t
d
T
=
−
K
(
2
T
1
+
T
2
−
T
0
)
⇒
4
30
=
−
K
(
2
170
−
20
)
...(i)
t
20
=
−
K
(
2
120
−
20
)
...(ii)
Divide
(
i
)
&
(
ii
)
we get
20
t
×
4
30
=
−
K
(
60
−
20
)
−
K
(
85
−
20
)
⇒
t
=
3
13
min