A cube of side 10 cm and unknown mass is suspended from one end of a uniform rigid rod. A 200 g mass is suspended from the other end. The support point is 2 cm from the cube end and 25 cm from the 200 g mass, as shown in the original figure. Water is added until half the cube volume is submerged and the system balances. Take water density = 1 g/cm³ and g = 10 m/s². Find the cube mass in kg.

Question

G1 Physics · Accepted Answer

Cube volume: V = (0.10)³ = 1×10⁻³ m³. Half the volume is submerged, so displaced water volume is V s u b V_sub V s ​ u b = 0.5×10⁻³ m³. Buoyant force: F B F_B F B ​ = ρ_water g V s u b V_sub V s ​ u b = 1000×10×0.5×10⁻³ = 5 N. Let the cube mass be m kg. Take moments about the support point O. The cube's effective downward force is mg - F B . F_B. F B ​ . Its lever arm is 2 cm = 0.02 m. The 0.2 kg mass has weight 0.2g = 2 N and lever arm 25 cm = 0.25 m. Torque balance: (mg - 5)(0.02) = 2(0.25).…