What is conserved in the following motion? Kinetic energy or angular momentum ? only one answer there is a body which is moving with a finite velocity and its velocity is perpendicular to its acceleration at any instant.
Now I'm confused as no torque is applied angular momentum must be conserved ,again kinetic energy is  a scalar quantity ,so this must be conserved too but any one of them have to be correct because it was a MCQ question asked in an exam . Which one will be more accurate ?
 A: Angular momentum is not necessarily conserved.
Since acceleration is perpendicular to velocity, we can change the object's position/direction but not its speed.
Recall angular momentum is $\vec L = \vec r \times \vec v$ and because the object's position is free to change "however we'd like," so is its angular momentum -- we can change $\vec r$ however we'd like, and the direction of $\vec v$, which means that $\vec L$ can change as well.
A: Since the acceleration($\frac{d\vec{v}}{dt}$) is perpendicular to the velocity($\vec{v}$) at any instant, we have $\frac{d\vec{v}}{dt} \cdot \vec{v} = 0$ at any instant. This implies $\frac{d}{dt}(\vec{v}\cdot \vec{v})=2\frac{d\vec{v}}{dt} \cdot \vec{v} = 0$. Since KE=$\frac{1}{2}m\vec{v}\cdot\vec{v}$ we must have $\frac{d}{dt}$KE=0 which means it's conserved. Another way to understand it is that since the net force is always perpendicular to the velocity, the work done by the net force is always zero and by the work-energy theorem the KE must remain unchanged. For the angular momentum, indeed the net torque may not be zero as $\frac{d\vec{L}}{dt}=\vec{r} \times m\dot{\vec{v}}$ is not zero in general.
