$A$ thin uniform rod of mass $M$ and length $L$ is rotating about a perpendicular axis passing through its centre with a constant angular velocity $\omega$. Two objects,each of mass $\frac{M}{3}$,are attached gently to the two ends of the rod. The rod will now rotate with an angular velocity of

If the Earth shrinks to $1/8$ of its original volume, while maintaining the same mass, then the duration of the day will be (in $hrs$)

Due to global warming, if the ice in the polar region melts and some of this water flows to the equatorial region, then

$A$ rod of mass $m$ and length $l$ is placed on a smooth floor and pivoted at its centre. $A$ gun of mass $m$ (without a shot) is mounted at one end. It fires a shot of mass $\frac{m}{10}$ with velocity $v$ relative to the gun. The angular velocity of the rod will be:

$A$ particle of mass $1 \ kg$ is subjected to a force which depends on the position as $\vec{F} = -k(x \hat{i} + y \hat{j}) \ N$ with $k = 1 \ kg \ s^{-2}$. At time $t = 0$,the particle's position is $\vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) \ m$ and its velocity is $\vec{v} = (-\sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{\pi} \hat{k}) \ m \ s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and $y$ components of the particle's velocity,respectively. Ignore gravity. When $z = 0.5 \ m$,the value of $(x v_y - y v_x)$ is . . . . . $m^2 \ s^{-1}$.

If the mass of the Earth is kept constant and its radius is reduced to half, then the duration of the day will become ........ $hr$.