$\mathrm{mg}+\mathrm{T}_{\mathrm{min}}=\frac{\mathrm{mv}_{\mathrm{min}}^{2}}{\mathrm{R}}$
$\frac{T_{\max }}{T_{\min }}=\frac{\frac{v_{\max }^{2}}{R}+g}{\frac{v_{\min }^{2}}{R}-g}$
$\Rightarrow \mathrm{v}_{\mathrm{m} \mathrm{in}}=10 \mathrm{m} / \mathrm{s}$
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$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is : $-\frac{m R}{M+m}$.
[$B$] The position of the point mass is : $x=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$.
[$C$] The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
[$D$] The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.
