Physics · 2012
NEET 2012 · Prelims · Code A · Q7
The potential energy of a particle in a force field is: U =A/(r^2)-B/r, where A and B are positive constants and r is the distance of particle from…
The potential energy of a particle in a force field is : $\displaystyle \mathrm{U}=\frac{\mathrm{A}}{\mathrm{r}^2}-\frac{\mathrm{B}}{\mathrm{r}}$,
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 :
Official answer
(4)
2A/B
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NEET (UG) 2012 Physics question, with the answer from NTA’s final answer key. Where our answers come from.