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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance \[{{C}_{0}}\]. Now one-third of the materialism replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area \[\frac{1}{3}\operatorname{A},\] dielectric constant 2 K and another with area \[\frac{2}{3}\operatorname{A},\]and dielectric constant K. If the capacitance of this new capacitor is C then \[C/{{C}_{0}}\]is:     JEE Main Online Paper ( Held On 25  April 2013 )

    A)                 \[1\]                                     

    B)                                        \[\frac{4}{3}\]

    C)                                        \[\frac{2}{3}\]                                   

    D)                                        \[\frac{1}{3}\]

    Correct Answer: B

    Solution :

                    \[{{C}_{0}}=\frac{k{{\in }_{0}}A}{d}\] \[C=\frac{k{{\in }_{0}}2}{3d}+\frac{2k{{\in }_{0}}A}{3d}=\frac{4}{3}\frac{k{{\in }_{0}}A}{d}\] \[\therefore \]\[\frac{C}{{{C}_{0}}}=\frac{\frac{4}{3}\frac{k{{\in }_{0}}A}{d}}{\frac{k{{\in }_{0}}A}{d}}=\frac{4}{3}\]


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