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AIIMS 2009
Exam
AIIMS
Year
2009
Subject
Physics
Chapter
Capacitance

AIIMS 2009 · Physics · Capacitance

A conducting sphere of radius R carrying charge Q lies inside an...

A conducting sphere of radius RR carrying charge QQ lies inside an uncharged conducting shell of radius 2R2 R. If they are joined by a metal wire, the amount of heat that will be produced is

Options

  1. A14πε0⋅Q24R\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 R}
  2. B14πε0⋅Q22R\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{2 R}
  3. C14πε0⋅Q2R\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{R}
  4. D24πε0⋅Q23R\frac{2}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{3 R}

Correct answer

A. 14πε0⋅Q24R\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 R}

Step-by-step solution

The capacitances of two are C1=4πε0RC_1=4 \pi \varepsilon_0 R and C2=4πε0(2R)C_2=4 \pi \varepsilon_0(2 R) Initial energy =Ei=Q22C1=E_i=\frac{Q^2}{2 C_1} Final energy =Ef=Q22C2=E_f=\frac{Q^2}{2 C_2} Heat produced =Ei−Ef=E_i-E_f=Q22[14πε0R−12×4πε0R]=14πε0⋅Q22R[1−12]=14πε0⋅Q24R\begin{aligned} & =\frac{Q^2}{2}\left[\frac{1}{4 \pi \varepsilon_0 R}-\frac{1}{2 \times 4 \pi \varepsilon_0 R}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{2 R}\left[1-\frac{1}{2}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 R}\end{aligned}$$
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