Calculate the work done in rotating a vertical meter scale of mass $400 \, g$ by an angle of ${60^o}$. (in $, J$)

$A$ particle is placed at the origin and a force $F = kx$ is acting on it (where $k$ is a positive constant). If $U(0) = 0$,the graph of $U(x)$ versus $x$ will be (where $U$ is the potential energy function):

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

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 $a = \dots \dots \dots \dots$

$A$ chain of mass $m$ and length $l$ is hanging freely from edge $A$ (as shown in diagram $I$). Calculate the work done to fold it as shown in diagram $(II)$.

The potential energy of a particle is given by $U = A - Bx^2$,where $x$ is the displacement. What is the magnitude of the force acting on the particle?