The potential energy of a particle in a certain field has the form `U=a/r^2-b/r`, where a and b are positive constants, r is the distance from the center of the field. Then

Options

A. At `r=(2a)/b`, particle is in steady equilibriumn.
B. At `r=(2a)/b` , particle is in unsteady equilibrium.
C. Maximum magnitude of force of attraction is `b^3/(27a^2)`
D. Maximum magnitude of force of attraction is `(27b^3)/a^2`

Answer

A, C

Explanation

`(dU)/(dr)=((-2a)/r^3+b/r^2)`
For `(dU)/(dr)=0, b/r^2=(2a)/r^3 rArr r=(2a)/b`
`(d^2U)/(dr^2)=(+6a)/r^4-(2b)/r^3=2/r^3((3a)/r-b)`
At `r=(2a)/b, (d^2U)/(dr^2)=2/r^3 ((3axxb)/(2a)-b)=2/r^3 b/2=b/r^3 gt 0`
i.e., U is minimum
So, it is a position of stable (steady) equilibrium.
`F=-(dU)/(dr)=(2a)/r^3-b/r^2`
For maximum force , `(dF)/(dr)=(-d^2U)/(dr^2)=0`
`rArr (-2)/r^3 ((3a)/r-b)=0 rArr r=(3a)/b`
`F=(2a)/((3a)/b)^3-b/((3a)/b)^2=(2ab^3)/(27a^3)-b^3/(9a^2)=-b^3/(27a^2)`