CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    If the energy released in the fission of one nucleus is 200 MeV. Then the number of nuclei required per second in a power plant of 16 kW will be:

    A)  \[0.5\times {{10}^{14}}\]                            

    B) \[0.5\times {{10}^{12}}\]

    C)  \[5\times {{10}^{12}}\]                

    D) \[5\times {{10}^{14}}\]

    Correct Answer: D

    Solution :

     Given: Energy released per second \[E=200\,MeV\]                                 \[=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\]                                 \[=3.2\times {{10}^{-11}}\,\text{joule}\] Power \[P=16\text{ }kW=16\times {{10}^{3}}\text{ watt}\] As the number of nuclei is given by                 \[n=\frac{P}{E}=\frac{16\times {{10}^{3}}}{3.2\times {{10}^{-11}}}=5\times {{10}^{14}}\]


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