A cube of side $10\ \text{cm}$ with unknown mass and a $200\ \text{g}$ mass are hung at the two ends of a light rigid rod $27\ \text{cm}$ long. The rod with the masses is placed on a wedge, the distance between the wedge point and the $200\ \text{g}$ mass being $25\ \text{cm}$. Initially the masses are not balanced. A beaker is placed beneath the unknown mass and water is added slowly to it. At a certain point the masses balance, with half the volume of the unknown mass inside the water. (Take the density of the unknown mass to be more than that of water, the mass does not absorb water, the water density is $1\ \text{g/cm}^3$ and $g = 10\ \text{m/s}^2$.) The unknown mass is ______ kg.
Numerical value type. Enter your answer.
Answer: 3
The cube hangs $27 - 25 = 2\ \text{cm}$ from the wedge. Half its volume ($500\ \text{cm}^3$) is under water, so the buoyant force is $F_B = 0.5\ \text{kg}\times10 = 5\ \text{N}$.
Torques about the wedge:
$$(mg - F_B)\times2 = 0.2\times10\times25 \Rightarrow 10m - 5 = 25 \Rightarrow m = 3\ \text{kg}$$
Solution by Sreeraj P, M.Sc Physics