RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    A black body is heated from \[{{27}^{o}}C\] to \[{{927}^{o}}C\] the ratio of radiations emitted will be:

    A)  1 : 256

    B)                                         1 : 64

    C)  1 : 16                    

    D)         1 : 4

    Correct Answer: A

    Solution :

    Here: Initial temperature \[{{T}_{1}}=27{}^\circ C=300K\]             Final temperature\[{{T}_{2}}=927{}^\circ C=1200K\] According to Stefans law that the radiant energy is                 \[E\propto {{T}^{4}}\,here\,\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{T_{1}^{4}}{T_{2}^{4}}\]             \[={{\left( \frac{300}{1200} \right)}^{4}}={{\left( \frac{1}{4} \right)}^{4}}=\frac{1}{256}\] Hence,       \[{{E}_{1}}:{{E}_{2}}=1:256\]


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