JEE Main & Advanced Sample Paper JEE Main - Mock Test - 30

  • question_answer
    In the figure shown, initially the spring of negligible mass is in undeformed state and the block has zero velocity, E is a uniform electric field. Then
    (i) The maximum speed of the block will be\[\frac{QE}{\sqrt{mK}}\]
    (ii) The maximum speed of the block will be \[\frac{2QE}{\sqrt{mK}}\].
    (iii) The maximum compression of the spring will be -\[\frac{QE}{K}\].
    (iv) The maximum compression of the spring will be \[\frac{2QE}{K}\].

    A) Only (i) and (iii) are correct                             

    B) Only (i) and (iv) are correct

    C) Only (ii) and (iii) are correct                            

    D) Only (ii) and (iv) are correct

    Correct Answer: B

    Solution :

    [b] For maximum compression, \[\frac{1}{2}K{{x}^{2}}=QEx\] \[{{x}_{\max }}=\frac{2QE}{K}\] \[{{V}_{\max }}=\frac{QE}{\sqrt{mK}}\]    (from conservation of energy)


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