A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water. The unknown mass is _____ kg.

The cube is attached at one end of a uniform rigid rod of total length $$27\text{ cm}$$. A wedge placed under the rod acts as a fulcrum. Given distance from the fulcrum to the $$200\text{ g}$$ mass is $$25\text{ cm}$$, the distance from the fulcrum to the cube is

$$27\text{ cm}-25\text{ cm}=2\text{ cm}$$.

Let the actual (true) mass of the cube be $$M\;(\text{kg})$$. Before the cube touches water, rotational equilibrium would require

$$Mg(2\text{ cm}) = 0.2g(25\text{ cm}) \; \Longrightarrow \; M=2.5\text{ kg}.$$

Since the rod is not balanced initially, the true mass must be greater than $$2.5\text{ kg}$$ (otherwise the $$200\text{ g}$$ side could never dominate).

A beaker is now placed beneath the cube and water is added until exactly half of the cube is submerged. By Archimedes’ principle, the upward buoyant force equals the weight of the displaced water.

Side of cube $$=10\text{ cm}=0.10\text{ m}$$. Total volume of cube: $$V = (0.10\text{ m})^{3}=1.0\times10^{-3}\text{ m}^{3}$$. Half the volume is submerged, so

$$V_{\text{disp}}=\frac{V}{2}=0.5\times10^{-3}\text{ m}^{3}.$$

With water density $$\rho_{w}=1000\text{ kg m}^{-3}$$, the buoyant force is

$$F_{B} = \rho_{w}gV_{\text{disp}} = 1000g(0.5\times10^{-3}) = 0.5g\;\text{N}.$$

This buoyant force is equivalent to the weight of a $$0.5\text{ kg}$$ mass. Hence the cube’s effective downward (apparent) weight becomes

$$\bigl(M-0.5\bigr)g.$$

At the moment of balance, clockwise and counter-clockwise moments about the fulcrum are equal:

$$\bigl(M-0.5\bigr)g(2\text{ cm}) = 0.2g(25\text{ cm}).$$

Cancel $$g$$ and substitute the lever arms (convert to the same units is optional since they cancel too):

$$\bigl(M-0.5\bigr)\times2 = 0.2\times25$$ $$\Longrightarrow \bigl(M-0.5\bigr)\times2 = 5$$ $$\Longrightarrow M-0.5 = \frac{5}{2} = 2.5$$ $$\Longrightarrow M = 2.5 + 0.5 = 3.0\text{ kg}.$$

Therefore, the mass of the cube is $$\mathbf{3\;kg}$$.