Force on the lower half of the rod by the upper half 
I broke the force into two components, along the rod as $T_1$ and perpendicular to the rod as $T_2$. If we see the forces on the lower half of the rod,
$$T_2 + \frac{mg}{2}\cos 60° = ma \\ T_2 = -\frac{mg}{4} + \alpha\frac{m}{2}\frac{3l}{4}; \qquad \alpha = \frac{3g}{4l} \\ T_2 = \frac{mg}{32}$$
But using torque equation, the value of $T_2$ comes out to be
$mg/16$ (which gives the answer).
I have a conceptual doubt. Is there some force which is not taken into consideration on the lower rod? Why do the forces differ when we solve using torque equation and force balancing?
 A: The principles behind this calculation are that

1. The resultant force on an isolated body equals its mass times the acceleration of its COM.
2. The resultant torque on an isolated body equals its moment of inertia about its COM times its angular acceleration (which is the same about all points).

The answer you have arrived at yourself ($T_2=\frac{1}{32}mg$) - from the calculation which you have shown using Principle #1- is correct. It correctly explains how the resultant force $F$ on portion CB causes the known acceleration $a_0$ of its COM  :
$$a_0=\alpha(\frac34L)=\frac{9}{16}g$$
$$F=T_2+\frac12mg\cos\theta=\frac{1}{32}mg+\frac14mg=\frac{9}{32}mg=(\frac12m)a_0$$
The answer you have been given ($T_2=\frac{1}{16}mg$) - and the calculation which you have NOT shown using Principle #2 - is incorrect. It does not provide a correct explanation for the known angular acceleration $\alpha$ of portion CB; neither does it explain the acceleration of the COM.
Your mistake is to assume that force $T_2$ provides the only torque on portion CB about its COM. Since this torque is anti-clockwise it cannot explain the clockwise rotation of portion CB about its COM. Force $T_2$ provides an anti-clockwise torque of $T_2(\frac14L)=\frac{1}{64}mgL$ whereas the rotation of CB requires a resultant clockwise torque of
$$I\alpha=\frac{1}{12}(\frac12m)(\frac12L)^2(3g/4L)=\frac{1}{128}mgL$$
In addition, portion AC must also exert a clockwise bending moment of $\frac{3}{128}mgL$ on portion CB, so that the resultant clockwise torque is $-\frac{1}{64}mgL+\frac{3}{128}mgL=+\frac{1}{128}mgL$.
Your missing calculation "using the torque equation" (ie Principle #2) has gone wrong because it does not model portion CB as a free body. By using the moment of inertia of CB about A instead of about its own COM you are inadvertently modelling the whole rod ACB as a rigid body, with CB rigidly connected to AC. The rigid body model assumes whatever internal forces and bending moments are required to maintain rigidity throughout the length of AB, even in the massless portion AC. The applied force $T_2$ acting on CB in this calculation reflects the missing mass of portion AC, which makes the solid rod AB rotate faster about A than the half-empty rod AB which has portion AC massless. Portion AC drags portion CB along with it.

Lessons :

*

*Always show your calculation in full. The act of explaining it to someone else forces you to question your logic and your assumptions and check your algebra, so that you often spot your own mistake. This practice gives you confidence that what you've done is correct. Failure to provide a full calculation suggests a lack of confidence in it.


*Do not assume that your textbook is correct. Answers to exercises
are often wrong, even if a skeleton solution is provided - because of typos, and because answers and skeleton solutions to exercises are not often checked during proof-reading. Full solutions in the text are almost always checked during proof-reading and are rarely wrong. Errors are more likely for obscure publishers than for well-known publishers.


*If you have checked your calculation several times and cannot see what you have done wrong, then you are probably correct. Have confidence in yourself. Far better to get it wrong and learn from your own mistake than to get it "right" because you have repeated someone else's mistake.
A: Your question sort of confuses me. If you cut the rod at point C, and focus on the rigid body BC, then the forces acting on BC would be its own weight, a tension force from rod AC, and a shear force from rod AC. There is also a bending moment acting on rod BC from rod AC.
What I would do if I were you is analyze the entire rod AB first. Can you find the acceleration of some point on that rod at this instance? If you can, then you can find the acceleration of any other point on rod AB, knowing the angular acceleration. One point you might be interested in is the CG of rod BC, which is 3L/4 from the pivot at A.
For angular quantities, note that first of all the angular velocity of the entire rod AB is zero at this instant immediately after the rope is cut. Can you find the angular acceleration at this instant? Additionally, note that rod AB has the same rotation, angular velocity, and angular acceleration of the cut rods AC and CB, since these sub-rods are material lines belonging to one rigid body: rod AB.
A: Consider the entire rod as a rigid body, and the net torque about point A; evaluate the angular acceleration of the rod.  The angular acceleration of the entire rigid rod is the angular acceleration of the portion CB. In polar coordinates, there is no motion of the rod in the radial direction.
Evaluate the linear acceleration of the center of mass of portion CB.  Do a force analysis on portion CB considering the force at point C from the portion AC, and the force of gravity: net force equals mass times acceleration of the center of mass (vector equation). Suggest using polar coordinates, and consider forces from portion AC in both radial and angular directions. From that force analysis evaluate the magnitude of the net force at point C on portion CB from the portion AC.
