Medium MCQ +4 / -1 PYQ · JEE Mains 2020

If the potential energy between two molecules is given by
U = $- {A \over {{r^6}}} + {B \over {{r^{12}}}}$,
then at equilibrium, separation between molecules, and the potential energy are :

  1. A ${\left( {{{2B} \over A}} \right)^{1/6}}$, $- {{{A^2}} \over {4B}}$ Correct answer
  2. B ${\left( {{{2B} \over A}} \right)^{1/6}}, - {{{A^2}} \over {2B}}$
  3. C ${\left( {{B \over A}} \right)^{1/6}},0$
  4. D ${\left( {{B \over {2A}}} \right)^{1/6}}, - {{{A^2}} \over {2B}}$

Solution

U = $- {A \over {{r^6}}} + {B \over {{r^{12}}}}$ <br><br>F = - ${{dU} \over {dr}}$ <br><br>= – (A(–6r<sup>–7</sup> )) + B(–12r<sup>–13</sup>) <br><br>for equilibrium, F = 0 <br><br>$\therefore$ 0 = ${{6A} \over {{r^7}}} - {{12B} \over {{r^{13}}}}$ <br><br>$\Rightarrow$ ${{6A} \over {12B}} = {1 \over {{r^6}}}$ <br><br>$\Rightarrow$ r = ${\left( {{{2B} \over A}} \right)^{{1 \over 6}}}$ <br><br>$\therefore$ U = $- {A \over {{{2B} \over A}}} + {B \over {{{\left( {{{2B} \over A}} \right)}^2}}}$ <br><br>= $- {{{A^2}} \over {2B}} + {{{A^2}} \over {4B}}$ <br><br>= $- {{{A^2}} \over {4B}}$

About this question

Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force

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