Keeping the mass of the Earth constant, if its radius is halved, then what will be the duration of the day (in $\text{ hours}$)?

$A$ 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 . . . . . $m^2 \ s^{-1}$.

$A$ thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity:

$A$ circular platform rotates in a horizontal plane about a vertical axis passing through its center. $A$ tortoise sits on the edge of the platform. The platform is given an initial angular velocity $\omega_0$. The tortoise now moves along a chord of the platform with a constant velocity (relative to the platform). How does the angular velocity $\omega(t)$ of the platform change with time $t$?

$A$ small disk is attached to the end of a light inextensible string,which passes through a hole in a frictionless horizontal tabletop. Initially,the disk moves on a circle of radius $R$ with kinetic energy $K_0$. The string is then slowly pulled so that the disk finally rotates on a circle of radius $\frac{R}{\eta}$. What is the work $W$ done in pulling the string?

If the volume of the Earth suddenly shrinks to $1/64$ of its present volume without any change in its mass,then the duration of the day on Earth will be ..... hours.