A vertical hollow cylinder is fixed on the ground. A uniform rod can be balanced partly in and partly out of the cylinder with the lower end of the rod resting against the vertical wall of the cylinder, as shown in the figure. The angle made by rod with the vertical in equilibrium is  $\theta$. Maximum and minimum possible value of  $\theta$ are   ${53}^0$ and  ${37}^0$, respectively. Coefficient of friction between road and cylinder is  $\mu =\tan \left[\frac{1}{2}{\tan }^{-1}\left(\frac{37}{28n}\right)\right]$.Find the value of n. 

Suppose the radius of the cylinder is R and length of rod is  $2I$.
Consider the case when the end A has sliding tendency up.
Forces acting on the rod are shown in the figure.
 
Resolving forces horizontally and vertically, we have
 $N_2=N_1\cos \alpha +\mu N_1\sin \alpha \mathrm{...}\text{(i)}$
and  $N_1\sin \alpha =\mu N_2\cos \alpha +w\mathrm{.....}\left(ii\right)$
Taking moment about A, 
 $N_1\left(2R\cos ec\alpha \right)=w\left(I\sin \alpha \right)\mathrm{...}\text{(iii)}$
From Eqs. (i), (ii) and (iii), we get 
 $\text{2R=I}[1-{\mu }^2]\sin \alpha -2\mu \cos \alpha ]{\sin }^2\alpha \mathrm{...}\text{(iv)}$
Similarly, when the rod makes least angle $\beta$ , we get A will have sliding tendency downward
 $\text{2R=I}[(1-{\mu }^2)\sin \beta +2\mu \cos \beta ]{\sin }^2\beta \mathrm{...}\text{(v)}$
From Eqs. (iv) and (v), we get
$\mu \text{ = tan}\left[\frac{1}{2}{\tan }^{-1}\left(\frac{{\sin }^3\alpha -{\sin }^3\beta }{{\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta }\right)\right]$  
After substituting the values, we get
$\mu \text{ = tan}\left[\frac{1}{2}{\tan }^{-1}\left(\frac{37}{84}\right)\right]\Rightarrow n=3$