$A$ particle moves along a circle with a constant angular speed $\omega$. Its displacement,with respect to the initial position of the particle at time $t = 0$,is plotted against time. The graph would look like:

$A$ body is projected at $t=0$ with a velocity $10 \ m/s$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t=1 \ s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g=10 \ m/s^2$,the value of $R$ is: (in $m$)

$A$ particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$. When the particle moves from one point to a diametrically opposite point,its

Particles $A$ and $B$ are moving with constant velocities along the $x$ and $y$ axes respectively. The graph of the separation $S$ between them with time $t$ is:

Two particles having mass $M$ and $m$ are moving in a circular path having radius $R$ and $r$. If their time periods are the same,then the ratio of their angular velocities will be:

$A$ particle of mass $1 \ kg$ is projected with an initial velocity $10 \ ms^{-1}$ at an angle of projection $45^{\circ}$ with the horizontal. The average torque acting on the projectile, between the time at which it is projected and the time at which it strikes the ground, about the point of projection in $Nm$ is: