JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A capacitor with capacitance \[5\mu F\]is charged to \[5\mu C.\]If the plates are pulled apart to reduce the capacitance to \[2\mu F,\]how much work is done? [JEE Main 9-4-2019 Morning]

    A) \[3.75\times {{10}^{6}}J\]  

    B) \[2.55\times {{10}^{6}}J\]

    C) \[2.16\times {{10}^{6}}J\]              

    D) \[6.25\times {{10}^{6}}J\]

    Correct Answer: A

    Solution :

    Work done \[=\Delta U\]             \[={{U}_{f}}-{{U}_{i}}\]             \[=\frac{{{q}^{2}}}{2{{C}_{f}}}-\frac{{{q}^{2}}}{2{{C}_{i}}}\]             \[=\frac{{{\left( 5\times {{10}^{-6}} \right)}^{2}}}{2}\left( \frac{1}{2\times {{10}^{-6}}}-\frac{1}{5\times {{10}^{-6}}} \right)\]             \[=\frac{15}{4}\times {{10}^{-6}}\]             \[=3.75\times {{10}^{-6}}J\] Option [a]

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