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NEET 2012 (AIPMT, Prelims) · Potential Energy & Equilibrium

Potential Energy & Equilibrium question from NEET 2012 (AIPMT, Prelims)

Q11NEET 2012 · AIPMT · PRELIMSMedium

Potential Energy & Equilibrium

The potential energy of a particle in a force field is U=Ar2−BrU = \frac{A}{r^2} - \frac{B}{r} where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

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Answer: (b)

  1. Equilibrium: dUdr=−2Ar3+Br2=0⇒r=2AB\frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} = 0 \Rightarrow r = \frac{2A}{B}
  2. d2Udr2=6Ar4−2Br3=1r3(3B−2B)>0\frac{d^2U}{dr^2} = \frac{6A}{r^4} - \frac{2B}{r^3} = \frac{1}{r^3}(3B - 2B) > 0 there, so it is stable.
  3. So the answer is (b) 2AB\frac{2A}{B}.
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