A) 32,000 K
B) 16,000 K
C) 8,000 K
D) 4,000 K
Correct Answer: D
Solution :
\[\frac{{{A}_{T}}}{{{A}_{2000}}}\,=\,\frac{16}{1}\,\,(given)\] Area under \[{{e}_{\lambda }}-\lambda \] curve represents the emissive power of body and emissive power \[\propto \,\,{{T}^{4}}\] \[(Hence\,\,area\,\,under\,{{e}_{\lambda }}-\lambda \,curve)\propto {{T}^{4}}\] \[\Rightarrow \,\,\,\,\frac{AT}{{{A}_{2}}_{000}}\,=\,\,{{\left( \frac{T}{2000} \right)}^{4}}\] \[\Rightarrow \,\,\frac{16}{1}\,\,=\,\,{{\left( \frac{T}{2000} \right)}^{4}}\,\,\,\Rightarrow \,\,T=4000\,\,K\]
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