Exams Net
Unrestricted Exams Practice
Home
Exams
Banking
CUET
Defence
Engineering
Finance
GATE
Insurance
International
JEE
LAW
MBA
MCA
Medical
Other
Police
PSC
RRB
SSC
State Govt
Subjectwise
Teacher
SET Exams
UPSC
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
CBSE
CBSE Class 10 Solutions
CBSE Class 12 Solutions
CBSE Question Papers (Pdf)
NCERT Books (Pdf)
NCERT Exemplar Books (Pdf)
NCERT Study Notes (Pdf)
CBSE Study Concepts (Pdf)
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Contact Us
Our Apps
Privacy
+
-
Test Index
Thermodynamics Part 3
Show Para
Hide Para
Section:
Physics
Share question:
© examsnet.com
Question : 10 of 100
Marks:
+1
,
-0
If
K
1
K_1
K
1
and
K
2
K_2
K
2
are the thermal conductivities,
L
1
L_1
L
1
and
L
2
L_2
L
2
are the lengths and
A
1
A_1
A
1
and
A
2
A_2
A
2
are the cross sectional areas of steel and copper rods respectively such that
K
2
K
1
=
9
,
A
1
A
2
=
2
,
L
1
L
2
=
2
\; \frac{K_2}{K_1}=9, \; \frac{A_1}{A_2}=2, \; \frac{L_1}{L_2}=2
K
1
K
2
=
9
,
A
2
A
1
=
2
,
L
2
L
1
=
2
. Then, for the arrangement as shown in the figure, the value of temperature
T
T
T
of the steel - copper junction in the steady state will be:
[27-Jul-2022-Shift-1]
1
8
∘
C
18^{\circ} \mathrm{C}
1
8
∘
C
1
4
∘
C
14^{\circ} \mathrm{C}
1
4
∘
C
4
5
∘
C
45^{\circ} \mathrm{C}
4
5
∘
C
15
0
∘
C
150^{\circ} \mathrm{C}
15
0
∘
C
Validate
Solution:
d
θ
d
t
=
K
1
A
1
l
1
(
T
1
−
T
)
=
K
2
A
2
l
2
(
T
−
T
2
)
\; \frac{d\theta}{dt} = \; \frac{K_1 A_1}{l_1} (T_1 - T) = \; \frac{K_2 A_2}{l_2} (T - T_2)
d
t
d
θ
=
l
1
K
1
A
1
(
T
1
−
T
)
=
l
2
K
2
A
2
(
T
−
T
2
)
⇒
450
−
T
T
−
0
=
K
2
A
2
l
1
K
1
A
1
l
2
=
9
×
1
2
×
2
\Rightarrow \; \frac{450 - T}{T - 0} = \; \frac{K_2 A_2 l_1}{K_1 A_1 l_2} = 9 \times \; \frac{1}{2} \times 2
⇒
T
−
0
450
−
T
=
K
1
A
1
l
2
K
2
A
2
l
1
=
9
×
2
1
×
2
⇒
450
−
T
=
9
T
⇒
T
=
4
5
∘
C
\Rightarrow 450 - T = 9 T \Rightarrow T = 45^{\circ} \mathrm{C}
⇒
450
−
T
=
9
T
⇒
T
=
4
5
∘
C
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
Prev Question
Next Question
More Free Exams
AIEEE Previous Papers
BITSAT Exam Previous Papers
JEE Adv
JEE Advanced Model Papers
JEE Advanced Previous Papers
JEE Main PYQ
JEE Mains Model Papers
VITEEE Previous Papers
