A particle of mass m is moving in a circular orbit of radius r in a force field given by `vec(F)=-(K)/(r^(2))hat(r)` . The angular momentum L of the particle about the centre varies as

Options

A. `Lpropsqrt(r)`
B. `Lpropr^(3//2)`
C. `Lprop(1)/(sqrt(r))`
D. `Lpropr^(1//3)`

Explanation

To solve the problem, we need to find how the angular momentum \( L \) of a particle of mass \( m \) moving in a circular orbit of radius \( r \) varies in a force field given by \( \vec{F} = -\frac{K}{r^2} \hat{r} \).

### Step-by-Step Solution:

1. **Understanding the Force**:
The force acting on the particle is given by \( \vec{F} = -\frac{K}{r^2} \hat{r} \). The negative sign indicates that the force is directed towards the center of the circular path, acting as a centripetal force.

2. **Centripetal Force Requirement**:
For an object moving in a circular path, the centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the linear speed of the particle.

3. **Equating Forces**:
Since the force provided by the field must equal the centripetal force required to keep the particle in circular motion, we can set them equal:
\[ \frac{mv^2}{r} = \frac{K}{r^2} \]

4. **Solving for Velocity**:
Rearranging the equation gives:
\[ mv^2 = \frac{K}{r} \]
Thus, we can solve for \( v^2 \):
\[ v^2 = \frac{K}{mr} \]

5. **Finding Angular Momentum**:
The angular momentum \( L \) of the particle about the center is given by:
\[ L = mvr \]
Substituting \( v \) from the previous step:
\[ L = m \left( \sqrt{\frac{K}{mr}} \right) r \]
Simplifying this expression:
\[ L = m \cdot r \cdot \sqrt{\frac{K}{mr}} = \sqrt{K \cdot m} \cdot \sqrt{r} \]

6. **Final Expression**:
Therefore, we can express the angular momentum as:
\[ L = \sqrt{K \cdot m} \cdot \sqrt{r} \]
This shows that the angular momentum \( L \) varies as \( \sqrt{r} \).

### Conclusion:
The angular momentum \( L \) of the particle about the center varies as \( L \propto \sqrt{r} \).