$A$ small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string,causing the radius of the circular motion to decrease by a factor of $2$,the kinetic energy of the mass will

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

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

The Sun rotates around its centre once in $27$ days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence (in $days$)? Assume the Sun to be a sphere of uniform density.

$A$ circular stage is free to rotate about a vertical axis passing through its centre. $A$ tortoise is sitting at the corner of the stage. The stage is provided with an initial angular velocity $\omega_0$. If the tortoise starts moving along one chord at a constant speed with respect to the stage,then how does the angular velocity of the stage $\omega(t)$ vary with time $t$?

$A$ man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is $75 \%$ of that with outstretched arms,then his rotational kinetic energy will