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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If temperature of the block is raised  by\[\Delta T,\] the potential difference V across the capacitor is :

    A) \[\frac{ms\Delta T}{C}\]

    B) \[\frac{mC\Delta T}{s}\]

    C) \[\sqrt{\frac{2mC\Delta T}{s}}\]

    D) \[\sqrt{\frac{2ms\Delta T}{C}}\]

    Correct Answer: D

    Solution :

                     \[E=\frac{1}{2}C{{V}^{2}}\]                         ?(i) The energy stored in capacitor is lost in form of heat energy. \[H=ms\,\Delta T\]                                    ...(ii) From Eqs. (i) and (ii), we have \[ms\,\Delta t=\frac{1}{2}C{{V}^{2}}\] \[V=\sqrt{\frac{2\,ms\,\Delta \Tau }{C}}\]


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