A particle moves with constant angular velocity in a circle. During the motion its
A. Energy is conserved
B. Momentum is conserved
C. Energy and momentum both are conserved
D. None of the above is conserved

Hint:In this question it is mentioned that a particle has constant angular velocity which means that the particle is carrying out uniform circular motion; so we will use rules which are applicable for Uniform circular motion.

Formula used:
The formula of kinetic energy and linear momentum is,
\[KE=\dfrac{1}{2}m{{v}^{2}}\] and \[\overset{\to }{\mathop{P}}\,=m\overset{\to }{\mathop{v}}\,\]
Here, \[\left( KE \right)\] represents kinetic energy of the particle, \[\left( m \right)\] represents mass of the particle, \[\left( v \right)\]represents speed of the particle, \[\left( \overset{\to }{\mathop{v}}\, \right)\] represent velocity of the particle and \[\left( \overset{\to }{\mathop{P}}\, \right)\] represent momentum of the particle.

Complete step by step solution:
The particle is carrying out uniform circular motion which means that it is moving with constant speed but not constant velocity as direction of velocity keeps on changing (because it is a vector quantity and vector quantities depend on direction as well as magnitude) and momentum is also a vector quantity as it is directly proportional to velocity therefore momentum cannot be conserved in this question.

As momentum is a vector quantity which is proportional to another vector quantity i.e. velocity so, momentum changes at every point. Therefore, option (B) and option (C) are not correct. We know that \[KE \propto {{v}^{2}}\] and we also know that speed is constant ( as the motion is uniform circular motion), which means that the kinetic energy is conserved.

Therefore, option (A) is the correct answer.

Note: Please make sure that in uniform circular motion speed is constant or conserved and not velocity (being a vector quantity changes at every point). Be very clear in differentiating scalar quantities from vector ones.