$ABC$ is an equilateral triangle with $O$ as its centre. $\vec F_1, \vec F_2$ and $\vec F_3$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about $O$ is zero,then the magnitude of $\vec F_3$ is

$A$ solid sphere of mass $M$ and radius $R$ has a moment of inertia $I$ about its diameter. It is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. The radius of the disc will be:

$A$ rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical,its lower half separates without exerting any reaction at the breaking point. Then the maximum angle '$\theta$' made by the hinged upper half with the vertical is ......... $^o$.

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_{0}$ in the $x-y$ plane as shown in the figure. For time $t < \frac{v_{0} \sin \theta}{g}$,the angular momentum of the particle is (where $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the $x, y$ and $z$ axes respectively):

$A$ simple pendulum of length $\ell$ is displaced so that its taut string is horizontal and then released. $A$ uniform bar of length $L$ pivoted at one end is simultaneously released from its horizontal position. If their motions are synchronous (i.e.,they have the same angular velocity at any angle $\theta$ below the horizontal), what is the length $L$ of the bar?

$A$ thin ring of mass $2 \ kg$ has a radius of $0.5 \ m$. It is rolling without slipping on a horizontal plane with a velocity of $1 \ m/s$. $A$ small ball of mass $0.1 \ kg$ moving in the opposite direction with a velocity of $20 \ m/s$ hits the ring at a height of $0.75 \ m$ and moves vertically upward with a velocity of $10 \ m/s$ after the collision. Immediately after the collision: