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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    A body radiate \[Q\text{ }cat/c{{m}^{2}}\]heat at a temperature\[{{227}^{o}}C\]. The rate of energy radiate at a temperature \[{{727}^{o}}C\]is (in\[cal/c{{m}^{2}}\]):

    A)  \[2Q\]                                 

    B)                         \[4Q\]                                 

    C)         \[16Q\]               

    D)         \[32Q\]

    Correct Answer: C

    Solution :

                    Rate of emission of radiation is proportional to the 4th power of temperature. \[Q\propto {{T}^{4}}\] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}\left( \frac{227+272}{727+273} \right)={{\left( \frac{500}{1000} \right)}^{4}}\]                 \[{{Q}_{2}}=16Q\]


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