 This is one of the doubts that is asked by one of our students in NPSS Shunpur class 12th. So if you don't know, this is the question which was asked in recently conducted JEE mains. So this is JEE mains that was conducted on January 9th. So you can read the question. So we have an L shaped object that is made up of thin rods of uniform mass density. So mass is uniformly distributed and it is suspended with a string as shown in the figure. So at one of the ends, the string is attached and AB is equal to BC. So angle theta is the angle made by the vertical with AB. We need to find what is the value of tan of theta. So in the options, you have various options with respect to what is the value of tan theta. So let us try to evaluate how we will go about it. So one thing is sure that this is an extended object which is not rotating. So this will be under rotational equilibrium as well as translational equilibrium. So when you apply the translational equilibrium equation which is net force is equal to 0, you will unnecessarily end up with another variable over here which is tension. So I don't want that tension to come in the equation. So what I will do, I will directly try to solve this using rotational equilibrium condition. So what I will do, I will balance all the torques that are applied at point A and equate that to 0. So then the torque due to the tension will be 0, isn't it? Now definitely this L-shaped rod will not be like this. Slightly it will tilt towards the right-hand side. So for the sake of better understanding, let us modify this entire scenario and let me draw it something like this that it will tilt a bit. So now let us say that it will be like this. It is hinged at this point, so this is A and this is my vertical. Now if this extended object, I can treat it as if it is made up of two parts A, B and B, C. Then you can say that at part AB, there is one force MG that is applied over here. So M is the mass of length A to B. Similarly M will be the mass of B to C length. Now you can see that this MG force will be applying force and will be trying to rotate it like this, entire structure. So there has to be a countering torque, let's say because of this rod B to C. That will be again MG only. So this MG should be applied from the center of mass of AB and this MG should be applied from center of mass of B to C. So I will extend this a bit so that it looks to be center of BC. Now there are no other forces apart from these two, isn't it? So I will try to balance a torque because of these two forces about this point. So let us assume that this is angle theta, torque because of this MG force will be what? MG into this perpendicular distance, isn't it? So I know what is this length. This length is L by 2 because this point is the center of mass. So this length, this one I am talking about is L by 2 sin theta. So torque due to AB part will be equal to MG into L by 2 sin theta. Now I have to find out torque because of this MG. For that I need to know how much is this perpendicular distance. What I know is that this distance is L by 2 because this MG should be acting from the center of mass and I need to find out what is this distance. So I will subtract these two distances. So just a small correction guys. This L by 2 is not the horizontal distance, the L by 2 is this distance along the length. So this is L by 2. So I just need to find out what is this length and subtract it from L by 2. Now you can see that this entire length A to B is L and this is angle theta. So this length let us say that is x. So x will be equal to what? It will be L. You can see that this is a triangle whose this length is L and this is the angle theta. This is the base. So L tan of theta will be this length x, isn't it? So you can see that this is 90 degree. This is length L. So x will be equal to L tan theta. So this length let me call it as y, y will be equal to L by 2 minus L tan of theta. Now the torque because of this MG where this torque because of BC length will be MG into L by 2 minus L tan theta which is this length and then I need to get this one. So L by 2 minus x is this length but I am interested in finding this perpendicular length. So if this is angle theta, this is angle theta, this will be 91 theta and this will be theta angle. So this angle will be theta. So this length will be equal to L by 2 minus L tan theta which is a habit in us that times cos theta. So when you find the torque because of MG you need to multiply MG with the perpendicular distance. So this is the habit in us that into cos of theta that is torque because of this length. So let me rearrange the terms. It will be MG into L by 2 cos theta minus MG L now tan theta into cos theta is sin theta. Now this torque should be equal to that torque then only torque will be balanced. So when you equate the torques you will have MG L by 2 sin theta is equal to MG L by 2 cos theta minus MG L sin theta. When you rearrange the terms you will get 3 by 2 MG L sin theta is equal to MG L cos theta by 2. Now from here on it just simple mathematics you will get tan theta to be equal to 1 by 3. So like this you have to solve this question. I hope you have understood it completely. In case you have any doubts please feel free to get in touch with me. Thank you.