JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    A particle having mass m and charge q moves along a line under the action of electric field \[E=\alpha -\beta x,\] where \[\alpha \] and \[\beta \] are positive constants and x is the distance from a point where the particle initially is at rest. Therefore for an observer moving with an acceleration \[\frac{q\alpha }{m},\]
    (i) The motion of particle is oscillatory                   
    (ii) The amplitude of particle is \[\frac{\alpha }{\beta }\]
    (iii) The mean position of particle is \[x=\frac{\alpha }{\beta }\]              
    (iv) Maximum acceleration of particle is \[\frac{q}{\beta }\alpha \]

    A) Only (i) is correct                                       

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

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

    D) (i), (ii), (iii) and (iv) are correct

    Correct Answer: D

    Solution :

    [d] Acceleration of particle \[{{a}_{P}}=\frac{qE}{m}=\frac{q\alpha }{m}-\frac{q\beta x}{m}\] Clearly the motion is oscillatory \[F=qE=0\] \[x=\frac{\alpha }{\beta }=\] Amplitude Maximum acceleration of particle w.r.t. observer is \[\frac{q\beta x}{m}=\frac{q.\beta }{m}\,\frac{\alpha }{\beta }=\frac{q.\alpha }{\beta }\]


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