JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A parallel plate capacitor has plates with area A and separation between it is d. A battery charges the plates to a potential difference\[{{V}_{0}}\]The battery is then disconnected and a dielectric slab of thickness d is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is

    A)  K                                

    B)  \[\frac{1}{K}\]

    C)  \[\frac{A}{{{d}^{2}}K}\]              

    D)  \[\frac{{{d}^{2}}K}{A}\]

    Correct Answer: A

    Solution :

     \[\Sigma mvr=\,({{l}_{system}})\omega \] Now, \[\Rightarrow \] As battery is disconnected, Q remains unaltered. So, \[mv\frac{l}{2}=\frac{(2m)\,{{(2l)}^{2}}}{12}\omega =\frac{2m(4{{l}^{2}})}{12}\omega \] \[\Rightarrow \]            \[\omega =\frac{3v}{4l}\]

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