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$\boxed{\text {Correct solution is B}}$ .

Solution involving physics :

Equilibrium will be reached once hydrostatic pressures on both sides become equal:

$\begin{array}{ccccc} &&& p_{1}=p_{2} \\ &&& \rho _{X}gh_{1}=\rho _{Y}gh_{2} \end{array}$

From here we can already see that if $\rho _{X}<\rho _{Y}$ , then must be $h_{1}>h_{2}$ . More precisely:

$\frac{h_{1}}{h_{2}}=\frac{\rho _{Y}}{\rho _{X}}$

This corresponds to picture $B$ .

Solution by intuition :

Level of liquid in two arms is the same only when the pressure on both arms is equal.

$A$ can't be because the level of $Y$ liquid didn't change at all. That suggests that liquid $X$ presses liquid $Y$ with the same force as air implying that it has the same density as air. Unless liquid $X$ is a super-light liquid, this is certainly not possible.

$C$ shows two arms have the same level, which suggest that they are the same liquids with same densities. Again, not true.

$D$ shows lower level in left than in the right arm. That implies that the portion of liquid $X$ presses harder than the same portion of the liquid $Y$ . That's a contradiction because something with lower density (and automatically less mass given some constant volume) cannot press harder than something with greater density.

$B$ is the only reasonable choice. Pouring liquid $X$ in one arm will certainly make liquid in the other arm go up, but it will always have lower level than the arm where $X$ is poured.

The total weight of each arm needs to be the same. Higher density liquid can achieve that weight with shorter length of the tube. So the level at the high density end will be lower. The answer is B. (The option A would only work if the lighter liquid were as light as air, which we are assuming is not happening.)

The weight of the left and right hand side must be of equal for the system to be in equilibrium. As the density of liquid X on the left hand side is less than liquid y, the combined liquid X and y on the right must have greater volume then than the single liquid y on the right hand side.

Consequently the solution b is correct.

Argument: The height in each arm would depend on the pressure bearing down on the surface of the liquid.

Demonstration: Imagine one end of the tube were closed off and the pressure in that part = 0 In that case the liquid will rise to the limit in that end of the tube as the air pressure from the other end bears down on the liquid on that side. This is essentially what happens when you drink liquids out of a straw.

Therefore, adding any fluid with density > air in one end will cause the liquid in the other end to rise

In A, Y left = Y right, given information above this is incorrect (see above)

In C X+ Y left =Y right, which implies density X = density Y, which is false

In D (assuming equal air pressure on both sides) X+Y left > Y right, which implies density X > density Y

At this point elimination has given us the correct answer, but for the sake of closure

In B, liq X has displaced somewhere the volume between air pressure and Y, therefore is the correct choice