1. Condition for circular motion

For a particle moving in a circle of radius $r$ under a central potential $U(r)=-\frac{C}{r}, \qquad C>0,$ the corresponding central force is $F(r)=-\frac{dU}{dr}.$

Compute it: $\frac{dU}{dr}=\frac{d}{dr}\left(-\frac{C}{r}\right)=\frac{C}{r^2}.$ Hence, $F(r)=-\frac{C}{r^2}.$ So the force is attractive, with magnitude $|F|=\frac{C}{r^2}.$

2. Use centripetal force relation

For circular motion, the required centripetal force is $\frac{mv^2}{r}.$ Since the attractive central force provides this, $\frac{mv^2}{r}=\frac{C}{r^2}.$

Multiply by $r$: $mv^2=\frac{C}{r}.$ Thus, $v^2=\frac{C}{mr}.$

Taking square root, $v=\sqrt{\frac{C}{mr}}.$

3. Interpret the graph

This gives $v \propto \frac{1}{\sqrt{r}}.$ Equivalently, $r=\frac{C}{mv^2} \propto \frac{1}{v^2}.$

So as $r$ increases, $v$ decreases nonlinearly. The $v$ vs $r$ graph is a decreasing curve of the form $1/\sqrt{r}$, not a straight line.

4. Correct option

Therefore, the correct radius–velocity graph is the one representing $v \propto r^{-1/2}.$ That corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.