KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    A solid metallic sphere of radius r is enclosed by a thin metallic shell of radius 2r, A charge q is given to the inner sphere- When the inner sphere is connected to the shell by a metal wire, the heat energy generated in it is given by

    A) \[\frac{{{q}^{2}}}{\pi {{\varepsilon }_{0}}r}\]

    B) \[\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}r}\]

    C) \[\frac{{{q}^{2}}}{8\pi {{\varepsilon }_{0}}r}\]

    D)  \[\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}\]

    Correct Answer: D

    Solution :

    the capacitance of inner sphere is \[{{C}_{1}}=4\pi {{\varepsilon }_{0}}r\]
    the capacitance of the outer shell is \[{{C}_{2}}=4\pi {{\varepsilon }_{0}}(2r)=8\pi {{\varepsilon }_{0}}r\]
    Before connection, the total energy is
    \[{{U}_{1}}=\frac{{{q}^{2}}}{2{{C}_{1}}}=\frac{{{q}^{2}}}{8\pi {{\varepsilon }_{0}}r}\]
    After connection, the entire charge q of the inner sphere is transferred to the outer shell. Hence, energy after connection is
    \[{{U}_{2}}=\frac{{{q}^{2}}}{2{{C}_{2}}}=\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}\]\[\therefore \]Heat generated = \[{{U}_{1}}-{{U}_{2}}\]
    \[=\frac{{{q}^{2}}}{8\pi {{\varepsilon }_{0}}r}-\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}=\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}\]

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