What should be the value of the refractive index n of a glass rod placed in air so that the light entering through flat surface of the rod does not cross the curved surface of the rod?

Using Snell’s law
$\begin{array}{l}n=\frac{\sin {\left(i\right)}_{\max }}{\sin {\left(r\right)}_{\max }}=\frac{\sin {90}^0}{\sin {\left(r\right)}_{\max }}\\ \because i_{\max }={90}^{0}Thus,sin{\left(r\right)}_{max}=\frac{1}{n}=\sin {\theta }_c\\ \Rightarrow {\left(r\right)}_{max}={\theta }_c\end{array}$

Now if the minimum value of r' is greater than the ${\theta }_c$  i.e. in all cases the total internal reflection     will take place at the glass air interface.
Therefore the necessary condition for T.I.R is
$r\text{'}\ge {\theta }_c or({90}^0-{\theta }_c)\ge {\theta }_c ----\left(i\right)$
Taking sin both sides
$\begin{array}{l}\sin ({90}^0-{\theta }_c)\ge \sin {\theta }_c\\ \Rightarrow \cos {\theta }_c\ge \sin {\theta }_{c}orcot{\theta }_c\ge 1 --- \left(ii\right)\end{array}$

From the triangle, $\mathrm{co}t{\theta }_c=\frac{\sqrt{n^2-1}}{1}$   

From eq.(ii) we get
  $\sqrt{n^2-1}\ge 1\therefore n\ge \sqrt{2}$