f 1×OB = f 1′×OA； F2×OB = F2′×OA？
From the above two formulas, we can get f 1'F2'= f 1 F2 = 3:4①.
ρ water VG+3f 1'= g =ρVG? ②
13 rho water VG+3f2' = g = rho VG ③
The density of cylinder ρ=3ρ water = 3× 1.0g/cm3 = 3g/cm3. So, A is wrong;
(2) The lever always keeps a horizontal balance, so the object under the pulley always keeps still, and the depth of another liquid is 2cm less than that of water.
Therefore, the depth of an object immersed in another liquid is 12cm-2cm= 10cm.
The volume of water is v water = 50cm2× 20cm-10cm2×12cm = 880cm3.
The volume of unknown liquid v = 50cm2×18cm-10cm2×10cm = 800cm3.
Because the mass is equal, ρ water v water = ρV, so the density of unknown liquid ρ liquid = ρ water v water v =1g/cm3× 880cm3800cm3 =1.1g/cm3 =1×103kg. So, c is wrong;
(3)F3=OAOB× 13(G-F floating) = 2×13 (3×103kg/m3×10n/kg×10×/kloc-.
(4) p = rho liquid GH =1./kloc-0 /×103kg/m3×10n/kg× 0.18m =1980pa; Therefore, d is correct.
So choose D.