$A$ horizontal platform is rotating with a uniform angular velocity about a vertical axis passing through its center. At some instant, a viscous liquid of mass $m$ is dropped at its center, which is free to spread and eventually falls off the edge. During this time interval, the angular velocity

$ABC$ 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

$A$ particle of mass $m$ is moving along the side of a square of side '$a$',with a uniform speed $v$ in the $x-y$ plane as shown in the figure. Which of the following statements is false for the angular momentum $\vec L$ about the origin?

Which one of the following four graphs best depicts the variation with $x$ of the moment of inertia $I$ of a uniform triangular lamina about an axis parallel to its base at a distance $x$ from it?

$A$ particle of mass $M=0.2 \ kg$ is initially at rest in the $xy$-plane at a point $(x=-l, y=-h)$,where $l=10 \ m$ and $h=1 \ m$. The particle is accelerated at time $t=0$ with a constant acceleration $a=10 \ m/s^2$ along the positive $x$-direction. Its angular momentum and torque with respect to the origin,in $SI$ units,are represented by $\vec{L}$ and $\vec{\tau}$,respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x, y$ and $z$-directions,respectively. If $\hat{k}=\hat{i} \times \hat{j}$,then which of the following statement$(s)$ is(are) correct?
$(A)$ The particle arrives at the point $(x=l, y=-h)$ at time $t=2 \ s$.
$(B)$ $\vec{\tau}=2 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(C)$ $\vec{L}=4 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(D)$ $\vec{\tau}=\hat{k}$ when the particle passes through the point $(x=0, y=-h)$.

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that: