On a certain planet which is a uniform solid sphere of mass M and Radius R, due to rotation about its polar axis, ratio of maximum to minimum values of apparent acceleration due to gravity at the surface is 3 : 2 

Gravitational force of attraction between planet (M) and satellite (m) can be depicted as  $F=\frac{GMm}{R(R+l)}$
Satellite is in pure rotation about centre of earth so $F=m{a}_{cm}$ can be applied for satellite
$\frac{GMm}{R(R+l)}=m(R+\frac{l}{2}){\omega }^2\mathrm{...................}\left(1\right)$

Further

$$\begin{gathered} \frac{g_{eq}}{g_{pole}}=\frac{2}{3}\Rightarrow \frac{g-R{\omega }^2}{g}=\frac{2}{3} \\ \Rightarrow {\omega }^2=\frac{g}{3R}\Rightarrow {\omega }^2=\frac{GM}{3R^3}\mathrm{......................}\left(2\right) \\ \left(1\right) \& \left(2\right) \Rightarrow R(R+l)(R+\frac{l}{2})=3R^3 \end{gathered}$$

Solving the quadratic in $l$, we get  $l=R$

C,D are points of transaction (C: near, D: far)
Tension at  $C=T_1$, tension at $D=T_2$ . Acceleration of all parts of Rod is same and  from  previous question can be calculated as $(R+\frac{l}{2}){\omega }^2=\frac{3R}{2}\frac{GM}{3\text{}{R}^3}=\frac{GM}{2R^2}$

So for F.B.D of AC,

$\frac{GMm/3}{R(R+\frac{l}{3})}-T_1=\frac{m}{3}\left(\frac{GM}{2R^2}\right)$

$\Rightarrow T_1=\frac{GMm}{3}(\frac{3}{4R^2}-\frac{1}{2R^2})=\frac{GMm}{12R^2}$

So for F.B.D of AD,

$$\begin{gathered} \frac{GM(2m/3)}{R(R+\frac{2l}{3})}-T_2=\left(\frac{2m}{3}\right)\left(\frac{GM}{2R^2}\right) \\ \Rightarrow T_2=\frac{GM\left(2m\right)}{3}[\frac{3}{5R^2}-\frac{1}{2R^2}]=\frac{GMm}{15R^2} \\ \therefore \frac{T_1}{T_2}=\frac{5}{4} \end{gathered}$$