As there is no external torque, angular momentum will remain constant. When the tortoise moves from $A$ to $C$, figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from $C$ to $B$, moment of inertia increases. Therefore, angular velocity decreases



If, $M$=mass of platform

$R=$radius of platform

$m=$mass of tortoise moving along the chord $AB$

$a=$perpendicular distance of $O$ from $AB$

Initial angular momentum, $I_1=m{R}^2+\frac{M{R}^2}{2}$

At any time $t$, let the tortoise reach $D$ moving with velocity $v$

$\therefore AD=vt$

$AC=\sqrt{R^2-a^2}$

As $DC=AC-AD=(\sqrt{R^2-a^2}-vt)$

$\therefore OD=r=a^2+{\left[\sqrt{R^2-a^2}-vt\right]}^2$

Angular momentum at time $t$

$I_2=m{r}^2+\frac{M{R}^2}{2}$

As angular momentum is conserved

$\therefore I_1{\mathrm{\omega }}_0=I_2\mathrm{\omega }\left(t\right)$

This shows that variation of $\mathrm{\omega }\left(t\right)$ with time is nonlinear. Choice