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

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is:

If a body of mass $200\, g$ falls from a height $200\, m$ and its total $P.E.$ is converted into $K.E.$ at the point of contact of the body with the earth's surface,then what is the decrease in $P.E.$ of the body at the contact? $(g = 10\, m/s^2)$ ............ $J$

$A$ number of identical cubical blocks of edge $a$ and mass $m$ are lying on a horizontal table. The work done on the blocks in arranging them in a column of height $(n + 1)a$ on the table is

Consider a one-dimensional motion of a particle with total energy $E$. There are four regions $A, B, C$ and $D$ in which the relation between potential energy $V$,kinetic energy $K$,and total energy $E$ is as given below:
Region $A: V > E$
Region $B: V < E$
Region $C: K < E$
Region $D: V > E$
State with reason in each case whether a particle can be found in the given region or not.

$A$ particle is released from rest at the point $x = a$ and moves along the $x$ axis subject to the potential energy function $U(x)$ shown. The particle: