NEET Sample Paper NEET Sample Test Paper-76

  • question_answer
    Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of energy radiated per second by the first sphere to that by the second is

    A)  \[1 :1\]                         

    B)  \[16:1\]

    C)  \[4:1\]              

    D)  \[1:9\]

    Correct Answer: A

    Solution :

    Energy radiated per second by a body which has surface area A at temperature T given by Stefan?s law is \[E =\sigma A{{T}^{4}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{{E}_{1}}}{{{E}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\,{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}={{\left( \frac{1}{4} \right)}^{2}}\,{{\left( \frac{4000}{2000} \right)}^{4}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{16}{16}=\frac{1}{1}=1:1\]


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