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(C) The angular momentum $L$ of the system (platform + tortoise) about the axis of rotation is conserved because there is no external torque acting on

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(C) The angular momentum $L$ of the system (platform + tortoise) about the axis of rotation is conserved because there is no external torque acting on the system.$L = I(t) \omega(t) = 	ext{constant}$Here, $I(t)$ is the moment of inertia of the system at time $t$.Let $R$ be the radius of the platform, $M$ its mass, and $m$ the mass of the tortoise.The moment of inertia of the platform is $I_p = \frac{1}{2}MR^2$.The tortoise moves along a chord. Let the distance of the chord from the center be $d$. At time $t$, the distance of the tortoise from the center is $r(t) = \sqrt{d^2 + (vt)^2}$, where $v$ is the constant speed of the tortoise relative to the platform.The total moment of inertia is $I(t) = I_p + mr(t)^2 = I_p + m(d^2 + v^2t^2)$.Since $L = I(t) \omega(t)$, we have $\omega(t) = \frac{L}{I_p + m(d^2 + v^2t^2)}$.As the tortoise moves from the edge towards the center (or vice versa), the distance $r(t)$ first decreases until it reaches the point closest to the center (at $t=0$ if the chord passes through the center, or at some $t$ if it doesn't) and then increases.When $r(t)$ is minimum, $I(t)$ is minimum, and therefore $\omega(t)$ is maximum.Conversely, when $r(t)$ is maximum (at the edges), $I(t)$ is maximum, and $\omega(t)$ is minimum.Since the tortoise starts at the edge, $r(t)$ decreases as it moves towards the center, causing $I(t)$ to decrease and $\omega(t)$ to increase. After passing the point closest to the center, $r(t)$ increases, causing $I(t)$ to increase and $\omega(t)$ to decrease.Thus, the angular velocity $\omega(t)$ increases and then decreases.

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