1. Given potential energy

   $V(r)=\frac{kr^2}{2}$

   The force is obtained from:

   $\vec F=-\frac{dV}{dr}\hat r$

   So,

   $\frac{dV}{dr}=kr$

   hence,

   $\vec F=-kr\hat r$

   This is an attractive central force of magnitude

   $F=kr$

2. Condition for circular motion

   For a circular orbit of radius $R$, the required centripetal force is

   $\frac{mv^2}{R}$

   This must be provided by the central force at $r=R$:

   $kR=\frac{mv^2}{R}$

   Therefore,

   $mv^2=kR^2$

   $v^2=\frac{k}{m}R^2$

   $v=\sqrt{\frac{k}{m}}\,R$

   So option B is correct and option A is incorrect.

3. Angular momentum

   For circular motion, velocity is perpendicular to radius vector, so

   $L=mvR$

   Substitute $v=\sqrt{\frac{k}{m}}R$:

   $L=mR\left(\sqrt{\frac{k}{m}}R\right)$

   $L=\sqrt{mk}\,R^2$

   So option C is correct and option D is incorrect.

4. Final evaluation of options

   • A: False
   • B: True
   • C: True
   • D: False

Therefore, the correct options are:

$\boxed{\text{B, C}}$

5. Comparison with stored answer

   Stored correct answer: B, C

   Derived answer: B, C

   They match exactly.