Consider the two statements related to circular motion in usual notations:
$A$. In uniform circular motion,$\vec{\omega}$,$\vec{v}$,and $\vec{a}$ are always mutually perpendicular.
$B$. In non-uniform circular motion,$\vec{\omega}$,$\vec{v}$,and $\vec{a}$ are always mutually perpendicular.

$A$ projectile is launched from the origin in the $xy$ plane ($x$ is the horizontal and $y$ is the vertically up direction) making an angle $\alpha$ from the $x$-axis. If its distance $r = \sqrt{x^2 + y^2}$ from the origin is plotted against $x$,the resulting curves show different behaviors for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure. For $\alpha_1$,$r(x)$ keeps increasing with $x$,while for $\alpha_2$,$r(x)$ increases and reaches a maximum,then decreases and goes through a minimum before increasing again. The switch between these two cases takes place at a critical angle $\alpha_c$ (where $\alpha_1 < \alpha_c < \alpha_2$). The value of $\alpha_c$ is (where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity).

$A$ particle of mass $m$ performs uniform circular motion of radius $r$ with linear speed $v$ under the application of force $F$. If $m$,$v$,and $r$ are all increased by $20 \%$,the necessary change in force required to maintain the particle in uniform circular motion is: (in $\%$)

Two particles,one at the centre of a circle of radius $R$,and another at a point $Q$ on the circle,start moving towards a point $P$ on the circle at the same time (see figure below). Both move with uniform velocities $\vec{V}_1$ and $\vec{V}_2$ respectively. They reach the point $P$ at the same time. If the angle between the velocities is $\theta$ and the angle subtended by $P$ and $Q$ at the centre is $\phi$ (as shown in the figure),then:

$A$ particle is moving in a circle:

Two bodies of mass $10 \, kg$ and $5 \, kg$ are moving in concentric orbits of radii $R$ and $r$ such that their periods are the same. What is the ratio of their centripetal accelerations?