A small particle of mass `m`, moves in such a way that the potential energy `U = ar^(3)`, where a is position constant and `r` is the distance of the particle from the origin. Assuming Rutherford's model of circular orbits, then relation between `K.E` and `P.E` of the particle is :

Options

A. `K.E = (3)/(2) U`
B. `K.E = U`
C. `K.E = 3U`
D. `K.E = (U)/(2)`

Explanation

To solve the problem, we need to establish the relationship between the kinetic energy (K.E) and potential energy (P.E) of a particle moving in a potential field defined by \( U = ar^3 \).

### Step-by-Step Solution:

1. **Identify the Potential Energy**:
The potential energy \( U \) of the particle is given by:
\[ U = ar^3 \]
where \( a \) is a constant and \( r \) is the distance from the origin.

2. **Calculate the Force**:
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dr} \]
Therefore, we differentiate \( U \):
\[ F = -\frac{d}{dr}(ar^3) = -3ar^2 \]

3. **Relate Force to Centripetal Force**:
According to Rutherford's model, the force acting on the particle provides the necessary centripetal force for circular motion. Thus, we can equate the magnitude of the force to the centripetal force:
\[ F = \frac{mv^2}{r} \]
Setting the two expressions for force equal gives:
\[ -3ar^2 = \frac{mv^2}{r} \]

4. **Solve for Kinetic Energy**:
Rearranging the equation to solve for \( mv^2 \):
\[ mv^2 = -3ar^3 \]
The kinetic energy \( K.E \) is given by:
\[ K.E = \frac{1}{2} mv^2 \]
Substituting for \( mv^2 \):
\[ K.E = \frac{1}{2}(-3ar^3) = -\frac{3}{2}ar^3 \]

5. **Relate K.E to P.E**:
We know from the potential energy expression that:
\[ U = ar^3 \]
Therefore, we can express the kinetic energy in terms of potential energy:
\[ K.E = -\frac{3}{2}U \]
However, since we are looking for a positive relationship, we can rewrite it as:
\[ K.E = \frac{3}{2}(-U) \]
Thus, the relationship between kinetic energy and potential energy is:
\[ K.E = \frac{3}{2} U \]

### Final Result:
The relation between the kinetic energy and potential energy of the particle is:
\[ K.E = \frac{3}{2} U \]