Potential energy of a particle varies with position $r$ as,$U = \left( \frac{\alpha}{r^4} - \frac{\beta}{r^5} \right) \text{ J}$,where $\alpha$ and $\beta$ are positive constants. The particle will be in equilibrium at $r = \dots$

The energy stored in a wound watch spring is:

$A$ particle which is constrained to move along the $x-$axis is subjected to a force in the same direction which varies with the distance $x$ of the particle from the origin as $F(x) = -kx + ax^3$. Here $k$ and $a$ are positive constants. For $x \ge 0$,the functional form of the potential energy $U(x)$ of the particle is

If the potential energy of a particle is given by $U = A - Bx^2$,then the force is proportional to which of the following?

The potential energy of a particle is related to the $x$ coordinate by the equation $U = x^2 - 2x$. The particle will be in stable equilibrium at:

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$,where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D = [U(x = \infty) - U_{\text{at equilibrium}}]$,then $D$ is: