$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)

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:

$A$ smooth chain of length $2 \,m$ is kept on a table such that its length of $60 \,cm$ hangs freely from the edge of the table. The total mass of the chain is $4 \,kg$. The work done in pulling the entire chain on the table is (Take, $g=10 \,m/s^2$) (in $\,J$)

The following figure shows the variation of potential energy $V(x)$ of a particle with distance $x$. The particle has

The diagrams represent the potential energy $U$ as a function of the inter-atomic distance $r$. Which diagram corresponds to stable molecules found in nature?

The potential energy $(U)$ of a diatomic molecule is a function dependent on $r$ (interatomic distance) as $U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3$,where $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be $\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$,where $a = \dots \dots \dots \dots$