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JEE Main 2016 · Physics · Electrostatics · medium

A solid conducting sphere of radius R carries charge +Q. It is surrounded by a thin concentric conducting spherical shell of radius 2R carrying…

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A solid conducting sphere of radius RR carries charge +Q+Q. It is surrounded by a thin concentric conducting spherical shell of radius 2R2R carrying charge −3Q-3Q. The sphere is then joined to the shell by a thin conducting wire. The heat produced in the wire is
  1. A. Q28πε0R\dfrac{Q^2}{8\pi\varepsilon_0R}
  2. B. Q232πε0R\dfrac{Q^2}{32\pi\varepsilon_0R}
  3. C. Q216πε0R\dfrac{Q^2}{16\pi\varepsilon_0R}
  4. D. Q24πε0R\dfrac{Q^2}{4\pi\varepsilon_0R}
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Answer: (C) Q216πε0R\dfrac{Q^2}{16\pi\varepsilon_0R}

Let k=14πε0k=\dfrac{1}{4\pi\varepsilon_0}. Initially the field exists between RR and 2R2R (due to QQ) and outside 2R2R (net charge −2Q-2Q). Energy U1=kQ22(1R−12R)+k(2Q)22(2R)=kQ24R+kQ2R=5kQ24RU_1=\dfrac{kQ^2}{2}\left(\dfrac1R-\dfrac{1}{2R}\right)+\dfrac{k(2Q)^2}{2(2R)}=\dfrac{kQ^2}{4R}+\dfrac{kQ^2}{R}=\dfrac{5kQ^2}{4R}. After connection all of the charge sits on the outer surface of the shell (net −2Q-2Q), the field inside vanishes, and U2=k(2Q)22(2R)=kQ2RU_2=\dfrac{k(2Q)^2}{2(2R)}=\dfrac{kQ^2}{R}. The energy lost as heat is U1−U2=kQ24R=Q216πε0RU_1-U_2=\dfrac{kQ^2}{4R}=\dfrac{Q^2}{16\pi\varepsilon_0R}.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake