Question

The given plot shows the variation of potential energy $U$ due to the interaction between two particles. Particles are separated by a distance $r$. Identify the correct statement out of the given statements:

$1$. $B$ and $D$ are equilibrium points.
$2$. $C$ is a point of stable equilibrium.
$3$. The force of interaction between the two particles is attractive between points $B$ and $C$, and repulsive between points $D$ and $E$ on the curve.
$4$. The force of interaction between the particles is repulsive between points $A$ and $C$.
Which of the above statements are correct?

• A. $1$ and $3$
• B. $1$ and $4$
• C. $2$ and $4$
• D. $2$ and $3$

Answer 1

The correct option is C $2$ and $4$


At point $C$, potential energy is minimum and $F=$ - Slope of $U-x$ curve $=0$
Hence, it is a point of stable equilibrium.
At $B$ and $D$, Slope of $U-x$ curve, $\frac{dU}{dx}\ne 0$. Hence, they are not equilibrium points.

From $A$ to $C$:
$F=-\frac{dU}{dx}=+ve$, $\because \text{slope}=\frac{dU}{dx}<0$
A $+ve$ force between interacting particles indicates the repulsion between particles and instability in that region of the curve.
Also from $B$ to $C$, slope of the curve is positive and which means force of interaction is repulsive.
Statements 2 and 4 are correct.
$\therefore$ Option $C$ is correct.