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JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 1 Questions
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© examsnet.com
Question : 32
Total: 100
Three rods of Copper, Brass and Steel are welded together to form a Y shaped structure. Area of cross - section of each rod = 4 cm
2
. End of copper rod is maintained at 100ºC where as ends of brass and steel are kept at 0ºC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:
[2014]
1.2 cal/s
2.4 cal/s
4.8 cal/s
6.0 cal/s
Validate
Solution:
Rate of heat flow is given by,
Q
=
K
A
(
θ
1
−
θ
2
)
l
Where, K = coefficient of thermal conductivity
l = length of rod and A = area of cross-section of rod
If the junction temperature is T, then
Q
Copper
=
Q
Brass
+
Q
Steel
0.92
×
4
(
100
−
T
)
46
=
0.26
×
4
×
(
T
−
0
)
13
+
0.12
×
4
×
(
T
−
0
)
12
⇒
2
00
−
2
T
=
2
T
+
T
⇒
T
=
40
°
C
∴
Q
Copper
=
0.92
×
4
×
60
46
=
4.8
c
a
l
∕
s
© examsnet.com
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