$A$ boy stands at the center of a horizontal platform with $2 \ kg$ masses in each hand,held close to his body. The platform rotates about a vertical axis passing through its center at a speed of $2 \ rev/s$. The moment of inertia of the system is $1 \ kg \cdot m^2$. When the boy extends his arms fully with the masses,the moment of inertia becomes $2 \ kg \cdot m^2$. In the second state,the kinetic energy of the system compared to the first state will be:

$A$ force $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$ is acting at a point $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$. The value of $\alpha$ for which angular momentum about the origin is conserved is

$A$ 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 radius of a rotating disc is suddenly halved while its mass remains the same,its angular velocity will become:

$A$ circular disc of mass $M$ has an initial angular velocity $\omega_1$. Two small spheres,each of mass $m$,are attached to the disc at opposite points on its diameter. What will be the final angular velocity of the disc?

$A$ solid sphere is rotating in space. If the radius of the sphere is increased while keeping its mass constant,which of the following physical quantities will not change?