Equilibrium of JEE!

I think all we need to do is that Centre of mass must lie in line of force.

Since mass density is constant, Mass of 2L=2M and mass of L=M, Now $2M(Lsin\alpha)=M\frac{Lcos\alpha}{2}\Rightarrow tan \alpha=0.25$ .

Here I have assumed $\alpha$ the angle between horizontal and 2L, so our answer will be 180-arctan 0.25=165.96 degrees!

Let $\lambda$ to be linear mass density so according to figure Figure ${ f }_{ 2 }\quad =\quad 2\mu \lambda Lg\\ \\ { f }_{ 1 }\quad =\quad \mu \lambda Lg\quad$ .

Now Rod is in equilibrium : $\because \quad { \tau }_{ o\quad ,\quad net }\quad =\quad 0$ . ${ f }_{ 2 }\quad L\sin { \alpha } \quad =\quad { f }_{ 1 }\quad \cfrac { L }{ 2 } \sin { (\alpha -\cfrac { \pi }{ 2 } ) } \\ \\ \alpha \quad =\quad \pi \quad -\quad \tan ^{ -1 }{ \cfrac { 1 }{ 4 } } \quad =\quad 165.96\quad degree$ .

Method 1

As the rod is moving with constant velocity, we can safely write the torque equation relative to any point on the rod (without bothering about the torque of pseudo force). The most convenient point is the right angled vertex.

$\lambda(2L)g(L\sin(\pi-\alpha))=\lambda Lg(L\cos(\pi-\alpha))\Rightarrow\tan\alpha=-\frac{1}{4}$

$\Rightarrow\alpha\approx166$ deg

Method 2

The centre of mass must lie along the line of the thread. Treating the two rods as point masses located at their respective CMs, we can say that the CM must divide the line joining them in the inverse ratio of masses.

$\frac{L\sin(\pi-\alpha)}{\frac{L}{2}\cos(\pi-\alpha)}=\frac{1}{2}\Rightarrow\alpha\approx166$ deg