So f float =G-F pull = 10N-5.2N=4.8N,
(2) Due to flooding, row V = object V =400cm3=4× 10-4m3.
From liquid gV with F = ρ:
ρ liquid =F floating gV row =4.8N 10N/kg×4× 10? 4m 3 = 1.2× 103kg/m3；
(3) liquid volume v =100cm2× 20cm = 2000cm3 = 2×10-3m3.
Liquid gravity g liquid =m liquid g=ρ liquid v liquid g =1.2×103kg/m3× 2×10-3m3×10n/kg = 24n,
The pressure of the container on the desktop F=(G liquid +G container +G object) -F tension = (24n+6n+10n)-5.2n = 34.8n;
Stress area: S= 100cm2= 10-2m2,
p=FS=34.8N 10？ 2m2=3480Pa。
Answer: (1) The buoyancy of an object immersed in liquid is 4.8n；;
(2) The liquid density in the cylinder is1.2×103 kg/m3;
(3) When the object is submerged, the pressure of the container on the desktop is 3480Pa. ..