NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer Two identical charged spheres suspended from a common point by two massless strings of lengths \[l,\]are initially at a distance \[d\,(d<<l)\]apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:                                                                      

    A)  \[v\propto {{x}^{\frac{1}{2}}}\]               

    B)  \[v\propto x\]

    C)   \[v\propto {{x}^{-\frac{1}{2}}}\]             

    D)   \[v\propto {{x}^{-1}}\]

    Correct Answer: C

    Solution :

                     \[\tan \theta =\frac{{{F}_{e}}}{mg}\simeq \theta \]                 \[\frac{K{{q}^{2}}}{{{x}^{2}}mg}=\frac{x}{2\ell }\]                 or            \[\]                                         ?..(1)                 or            \[{{x}^{3/2}}\propto q\]                                               ?..(2) differentiate eq.(i) w. r .t. time \[3{{x}^{2}}\frac{dx}{dt}\propto 2q\frac{dq}{dt}\]but \[\frac{dq}{dt}\] is constant so x\[{{x}^{2}}(v)\propto q\]      replace q from eq. (2) \[{{x}^{2}}(v)\propto {{x}^{3/2}}\] or \[\]

adversite


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