A solid cube of side 0.15 m has an average density of 210 kg m^–3.

(i) Calculate the weight of the cube.

(ii) The cube is placed in gasoline of density 720 kg m^–3. Calculate the proportion of the volume of the cube that is above the surface of the gasoline.

Water flows through a constricted pipe. Vertical tubes A and B, open to the air, are located along the pipe.

Describe why tube B has a lower water level than tube A.

i
F_weight = «ρgV_cube = 210×9.81×0.15³ =» 6.95«N»

ii
F_buoyancy = 6.95 = ρgV gives V = 9.8×10⁻⁴

\(\frac{{9.8 \times {{10}^{ - 4}}}}{{{{\left( {0.15} \right)}^3}}}\)=0.29 so 0.71 or 71% of the cube is above the gasoline

Award [2] for a bald correct answer.

«from continuity equation» v is greater at B
OR
area at B is smaller thus «from continuity equation» velocity at B is greater

increase in speed leads to reduction in pressure «through Bernoulli effect»

pressure related to height of column
OR
p=\(\rho \)gh