When the depth of the poured liquid h=4cm=0.04m, the pressure of the block at the bottom of the container f ′ =15n, while f ′ = g-f floats.
∴F float =G-F'=20N- 15N=5N,
∫F float =ρ liquid v row g=ρ liquid Shg,
∴ density of liquid:
ρ liquid =F floating SHG = 5n10x10x10? 4 m2×0.04m× 10N/kg = 1.25× 103kg/m3;
(2)∫G = mg =ρVg,
Density of block:
ρobject = GVG = 20n 10× 10× 10× 10? 6 m3× 10N/kg = 2× 103kg/m3,
∵ ρ substance >ρliquid,
When the depth of the poured liquid is h ′ =12 cm, the wood block will be immersed in the liquid and sink to the bottom of the container.
Pressure at the bottom of the container:
F pressure =G-F floating' =20N-ρ liquid VG = 20n-1.25x103kg/m3×10x10x.
Therefore, the answer is:1.25x103kg/m3; 7.5N。