A sphere of mass $2.8 \mathrm{kg}$ is rolling without slipping on an inclined plane of angle $37°$ from horizontal. Choose the correct statements(s).

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Angular} \mathrm{acceleration} \\ \frac{{\mathrm{R\mu }}_{\min }\mathrm{mg} \cos \mathrm{\theta }}{{\displaystyle \frac{2}{5}{\mathrm{mR}}^2}}=\frac{\mathrm{Rmg} \sin \mathrm{\theta }}{{\displaystyle \frac{7}{5}{\mathrm{mR}}^2}} \\ {\mathrm{\mu }}_{\min }=\frac{2}{7}\tan \mathrm{\theta }=\frac{2}{7}\times \frac{3}{4}=\frac{3}{14} \\ \left(\mathrm{ii}\right) {\mathrm{KE}}_{\mathrm{translation}}=\frac{1}{2}m{v}^2 \\ {\mathrm{KE}}_{\mathrm{rotation}}=\frac{1}{2}\times \frac{2}{5}{\mathrm{mR}}^2{\left(\frac{\mathrm{v}}{\mathrm{R}}\right)}^2=\frac{1}{5}{\mathrm{mv}}^2 \\ \left(\mathrm{iii}\right) \frac{{\mathrm{RF}}_{\mathrm{r}}}{{\displaystyle \frac{2}{5}{\mathrm{mR}}^2}}=\frac{\mathrm{Rmg} \sin \mathrm{\theta }}{{\displaystyle \frac{7}{5} {\mathrm{mR}}^2}} \\ {\mathrm{F}}_{\mathrm{r}}=\frac{2}{7}\mathrm{mg} \sin \mathrm{\theta }=\frac{2}{7}\times 2.8\times 10\times \frac{3}{5}=4.8 \mathrm{N} \\ \left(\mathrm{iv}\right) \left(\frac{\mathrm{Rmg} \sin \mathrm{\theta }}{\left(7/5\right){\mathrm{mR}}^2}\right)\mathrm{R}=\frac{5}{7}\mathrm{g} \sin \mathrm{\theta }=\frac{5}{7}\times 10\times \frac{3}{5}=\frac{30}{7} m/s^2 \end{gathered}$$