Properties of Matter Part 3

The cube side is $10\ \text{cm}$, so its volume is $V=10^3=1000\ \text{cm}^3$.Half of the cube is submerged, so the displaced water volume is $V_{\text{sub}}=500\ \text{cm}^3$.The mass of displaced water, and hence the buoyant-force equivalent mass, is $500\ \text{gm}=0.5\ \text{kg}$.At balance, the tension on the unknown-mass side is $T=(M-0.5)\ \text{kg}\cdot g$.The wedge is $25\ \text{cm}$ from the $200\ \text{gm}$ mass and $27-25=2\ \text{cm}$ from the unknown mass.Taking torques about the wedge: $(25\ \text{cm})(0.2\ \text{kg})g=(2\ \text{cm})(M-0.5)\ \text{kg}\cdot g$.Cancelling the common factors gives $25\times0.2=2(M-0.5)$.So $5=2M-1$.Hence $2M=6$, giving $M=3\ \text{kg}$.Therefore, the unknown mass is $\boxed{3\ \text{kg}}$.