A cube floats in water with 1/3rd part is outside the surface of water and it floats in another liquid with 3/4th part is outside the liquid then the density of liquid isSelect an answerA8/3B2/3C4/3D5/3

The problem is essentially asking us to find the density of the unknown liquid.

Step 1: Understand the problem The cube floats in water with 1/3rd of it above the surface. This means that 2/3rd of the cube is submerged in the water. Since the cube is in equilibrium, the weight of the cube is equal to the buoyant force of the water.

The cube also floats in another liquid with 3/4th of it above the surface. This means that 1/4th of the cube is submerged in the liquid. Again, since the cube is in equilibrium, the weight of the cube is equal to the buoyant force of the liquid.

Step 2: Apply Archimedes' principle Archimedes' principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.

Step 3: Set up the equation Let's denote the density of the cube as ρc, the density of water as ρw (which is 1 g/cm³), and the density of the unknown liquid as ρl.

From the first scenario, we have: ρc * V = ρw * (2/3)V ----(1)

From the second scenario, we have: ρc * V = ρl * (1/4)V ----(2)

Step 4: Solve the equation From equation (1), we can solve for ρc: ρc = ρw * (2/3) = 2/3 g/cm³

Substitute ρc into equation (2): 2/3 = ρl * (1/4) Solve for ρl: ρl = (2/3) / (1/4) = 8/3 g/cm³

So, the density of the unknown liquid is 8/3 g/cm³. Therefore, the answer is A: 8/3.