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NEET Sample Paper NEET Sample Test Paper-20

  • question_answer
    The binding energy of two nuclei\[{{P}^{n}}\] and \[{{Q}^{2n}}\]are \[x-\]joule and \[y-\]joule, respectively. Then the energy released in the reaction \[{{P}^{n}}+{{P}^{n}}={{Q}^{2n}}\]will be

    A) \[2x+y\]                      

    B) \[y-2x\]

    C)      \[xy\]               

    D)      \[x+y\]                        

    Correct Answer: B

    Solution :

    Energy released =final B.E.-initial B.E \[=y-2x\] Hence, the correction option is (b).


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