Tension In circular motion In the below question is given 2 scenarios in which we have to find the tension in two particular cases, A ball is held at rest in position A by two light cords.the horizontal cord is now cut  and the ball swings to the position B . what is the ratio of the tension in the cord in position B to that in position A ?

Now I can understand what the solutions says at A but at point B, it balances $T=Mgcosx$ for some reason, now shouldn't it be $Tcosx=mg$ because when u balance T in that direction at point B there is also a centrifugal force outwards which is not perpendicular to this, whilst when we balance the vertical component of tension, the centrifugal force is perpendicular to it and hence has no components.
 A: I may be misunderstanding your question but I still may be able to assist so here goes, and this is my first ever answer on this site, so please forgive me if I either tell you how to suck eggs, or express things in terms unfamiliar.
Here goes:
When the ball moves to the right and reaches zero velocity, there are two forces at work, and the system is not in equilibrium.
One force is due to gravity = $Mg$ (straight down).
The other force is due to the string. By virtue of the definition of a string is that the length of the string cannot change, then when working the vector diagram there can be no component of the resultant force in the direction of the string, i.e. the resultant force must be purely perpendicular (at right angles) to the string. This is how the ball will move, again, the system is not in equilibrium.
As such, the resultant force and the string tension force must form a right angle triangle, with the Gravity force Mg as the hypotenuse. From there you can see that the relationship between the Gravity force Mg and the string Tension T are defined by their adjacency to angle $x$ in the form $ T=Mg cos (x)$.
To come at this from the other angle, what if the answer were indeed  $ T cos (x) = Mg$?
This would mean on our free body diagram the spring tension would instead be the hypotenuse, with gravity and the resultant force (accelerating the ball) as the right angle triangles.
If you were to draw this, you would quickly notice that the resultant force would have a component parallel to the string. As such the resultant acceleration of the ball act to slacken the string, even though gravity was the force making it accelerate - this is a contradiction obviously.
I hope this has helped in some way; happy to answer any further questions.
