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AP EAMCET Engineering 2015 Code C Paper
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© examsnet.com
Question : 114
Total: 160
The ratio of energies of emitted radiation by a black body at 600k and 900k when the surrounding temperature is 300k
5
16
7
16
3
16
9
16
Validate
Solution:
E
=
σ
A
(
T
4
−
T
0
4
)
E
1
E
2
=
T
1
4
−
T
0
4
T
2
4
−
T
0
4
=
600
4
−
300
4
900
4
−
300
4
=
6
4
−
3
4
9
4
−
3
4
=
1296
−
81
6561
−
81
=
1215
6470
≈
3
16
© examsnet.com
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