$A$ stretched rubber band has:

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:

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

$A$ mass of $M \ kg$ is suspended by a weightless string of length $\ell$. The horizontal force required to displace it until the string makes an angle of $45^\circ$ with the initial vertical direction is:

$A$ uniform chain has a mass $m$ and length $l$. It is held on a frictionless table with one-sixth of its length hanging over the edge. The work done in just pulling the hanging part back on the table is:

The potential energy of a particle of mass $5 \ kg$ moving in the $XY$ plane is given by $V = -7x + 24y$ joules,where $x$ and $y$ are in metres. Initially at $t = 0$,the particle is at the origin $(0, 0)$ moving with a velocity of $\vec{v}_0 = 6[0.24 \hat{i} + 0.7 \hat{j}] \ m/s = [1.44 \hat{i} + 4.2 \hat{j}] \ m/s$. Then: