(C) The equilibrium position occurs where the force $F = -\frac{dU}{dr} = 0$.Given $U = A r^{-10} - B r^{-5}$.Differentiating with respect to $r$: $\f

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$\left(\frac{2A}{B}\right)^{\frac{1}{5}}$

(C) The equilibrium position occurs where the force $F = -\frac{dU}{dr} = 0$.Given $U = A r^{-10} - B r^{-5}$.Differentiating with respect to $r$: $\frac{dU}{dr} = -10 A r^{-11} + 5 B r^{-6}$.Setting $\frac{dU}{dr} = 0$ for equilibrium:$-10 A r^{-11} + 5 B r^{-6} = 0$.$5 B r^{-6} = 10 A r^{-11}$.$r^5 = \frac{10A}{5B} = \frac{2A}{B}$.Therefore,the equilibrium distance is $r = \left(\frac{2A}{B}\right)^{\frac{1}{5}}$.

Class 11 Physics Work, Energy, Power and Collision Potential Energy

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Potential energy as a function of $r$ is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$,where $r$ is the interatomic distance,and $A$ and $B$ are positive constants. The equilibrium distance between the two atoms will be

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A
$\left(\frac{A}{B}\right)^{\frac{1}{5}}$
B
$\left(\frac{B}{A}\right)^{\frac{1}{5}}$
C
$\left(\frac{2A}{B}\right)^{\frac{1}{5}}$
D
$\left(\frac{B}{2A}\right)^{\frac{1}{5}}$

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Potential energy as a function of $r$ is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$,where $r$ is the interatomic distance,and $A$ and $B$ are positive constants. The equilibrium distance between the two atoms will be

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In the equation $W = -mgh$,what does the negative sign indicate?

If the potential energy between two molecules is given by $U = -\frac{A}{r^6} + \frac{B}{r^{12}}$,then at equilibrium,the separation between the molecules and the potential energy are:

$A$ particle moves under the influence of a force $F = -2x$ in one dimension (where $x$ is the distance of the particle from the origin). Assume that the potential energy of the particle at the origin is zero. The schematic diagram of the potential energy $U$ as a function of $x$ is given by:

$A$ carpet of mass $M$ made of a material is rolled along its length in the form of a cylinder of radius $R$ and kept on a rough floor. If the carpet is unrolled without sliding to a radius $R/2$,the change in potential energy is ($g=$ acceleration due to gravity).

$A$ 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:

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