A body slides down the track from point P which is at a height of 10 cm. Maximum value of R for the body to successfully complete the loop is 
 

$$\begin{gathered} \mathrm{mg}(h-2R)=\frac{1}{2}m{v}^2 where v is the speed of the block at the highest point of the loop. \\ \Rightarrow \frac{m{v}^2}{R}=\frac{2mg(h-2R)}{R} . \\ The object will not loose contact with the loop at the highest point , \\ if \frac{m{v}^2}{R}\ge mg \Rightarrow \frac{2mg(h-2R)}{R}\ge mg \Rightarrow h\ge 2.5R \\ \Rightarrow R_{max}=\frac{10}{2.5} cm = 4 cm \end{gathered}$$