| Direction: In the following questions, more than one of one answers given are correct. Select the correct answers and mark it according to the following codes: |
| Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the some. The two bodies emit total radian power at the same rate. The wavelength \[{{\lambda }_{B}}\] corresponding to maximum spectral radiancy in the radiation from B is shifted from the wa9.-velength corresponding to maximum spectral radiancy in the radiation from A, by \[1.00\,\mu m\]. If the temperature of A is 5802 K |
| (1) The temperature of B is 1934 K |
| (2) \[{{\lambda }_{B}}=1.5\,\mu m\] |
| (3) The temperature of B is 11604 K |
| (4) The temperature of B is 2901 K |
A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: B
Solution :
[b] According to Stefans law \[E={{e}_{e}}A\sigma {{T}^{4}}\Rightarrow {{E}_{1}}={{e}_{1}}A\sigma T_{1}^{4}\,\,\text{and}\,\,{{E}_{2}}={{e}_{2}}A\sigma {{T}_{2}}^{4}\] \[\therefore \] \[{{E}_{1}}={{E}_{2}}\therefore {{e}_{1}}{{T}_{1}}^{4}={{e}_{2}}{{T}_{2}}^{4}\] \[\Rightarrow \] \[{{T}_{2}}={{\left( \frac{{{e}_{1}}}{{{e}_{2}}}{{T}_{1}}^{4} \right)}^{\frac{1}{4}}}={{\left( \frac{1}{81}\times {{\left( 5802 \right)}^{4}} \right)}^{\frac{1}{4}}}\] \[\Rightarrow \] \[{{T}_{B}}=1934K\] And, from Weiri's law \[{{\lambda }_{A}}\times {{T}_{A}}={{\lambda }_{B}}\times {{T}_{B}}\] \[\Rightarrow \]\[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{T{{ & }_{B}}}{{{T}_{A}}}\Rightarrow \frac{{{\lambda }_{B}}-{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{{{T}_{A}}-{{T}_{B}}}{{{T}_{A}}}\] \[\Rightarrow \]\[\frac{1}{{{\lambda }_{B}}}=\frac{5802-1934}{5802}=\frac{3968}{5802}\Rightarrow {{\lambda }_{B}}=1.5\mu m\]
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