Minimum radius for cylinder in rotation A rod of length 'l' and cylinder of radius 'r' is kept on an incline plane
as shown in the figure. The rod is pivoted, while the cylinder can roll
without slipping. A light string PQ attaches top of cylinder to some
point on the rod such that it is parallel to the incline. The minimum
value of radius of cylinder to ensure that the string is taut when the
system is released from rest is

The radius is to be found in terms on L.
The string will stay taut if at the time of release point q is moving faster than point p. Therefore acceleration of point q is more than p as it is released from rest.
FOR POINT Q: Torque about pivot $= \frac{L}{2}mgsinθ-2RF = \frac{mL^2}{3}α$ where f is the tension.
I assumed that the mass of both the cylinder and the rod is the same.
For point p: torque about bottom most point $= F 2R + mgsinθ R = \frac{3mR^2}{4}$
On equating torque Q > torque P i get a quadratic equation which cannot be simplified further.
The correct answer is $R > \frac{4L}{9}$.
How is this possible ? What am I missing or doing wrong ?
 A: Your mistake lies in considering the torques due to the string. We know that if the string is taut, their accelerations are going to be the same anyways. We need to find a value of $R$ such that the acceleration of $Q$ is greater than the acceleration of $P$ without the string, so than when the string is present, it remains taut.
Torque equation about base for P:
$ \frac {3 m_1 R^2}{2} \alpha_1 = 2m_1gR \sin \theta$
$\therefore a_1 = 2R \alpha _1 = \frac {4g \sin \theta}{3}$
Torque equation about base for Q:
$\frac {m_2 l^2}{3} \alpha _2 = \frac {m_2 gl \sin \theta}{2}$
$\therefore a_2 = 2R \alpha _2 = \frac {3gR \sin \theta}{l}$
Then solving the condition $a_2 > a_1$, we get the answer $ R > \frac {4l}{9}$
Note: You can also use the torque equations you have written (which include tension of the string). Get an expression for the tensile force by equating the linear accelerations of point $P$ and $Q$. Then apply the condition for string to remain taut, i.e. $F > 0$ and you should get the answer. However this method has unnecessary calculations, and is more difficult to solve, so I would recommend going by the method used in my answer.
A: For P and Q to have the same linear acceleration, you don't want to equate the torques. You need to equate the two α's.
