Question

A cubical block of wood, 10.7 cm on a side, floats at the interface between oil and water with its lower surface 2.10 cm below the interface (the figure ). The densityof the oil is 790 kg/m^3.

a) What is the gauge pressure at the upper face of the block?
? Pa

b)What is the gauge pressure at the lower face of the block?
? Pa

c)What is the mass of the block?
? kg

d)What is the density of the block?
? kg/m^3

Answer 1

a)
0 Pa

b)
P = ρ₁gh₁ +ρ₂gh₂
= (790)(9.8)(0.107) + (1000)(9.8)(0.021)
= 1034.194 Pa

c)

=((790)(0.107) + (1000)(0.021))/0.107
=986.26 kg/m³

Total volume of the object:
V = 0.107³ = 0.001225 m³

Mass:
m = (0.001225)(986.26)
= 1.208 kg

Answer 2

length of the side of the cubical block L = 10.7 cm= 0.107 mdensity of the oil ρ_o = 790 kg/m³density of thewater ρ_w = 790 kg/m³height h = 2.10 cm= 0.021 ma)the gauge pressure at the upper face of the block isP =ρ_ogh=(790 kg/m³)(9.8 m/s²)(0.021 m)= 162.582 Pab)the gauge pressure at the lower face of the block isP = ρ_oilgL + ρ_wgh= (790 kg/m³)(9.8 m/s²)(0.107 m) + (1000 kg/m³)(9.8 m/s²)(0.021 m)= 1034.194 Pac)mass of the block m = weight / g (since, weight = pressur*area)= PA/g= {(1034.194 Pa - 162.582 Pa) (0.107 m)²}/(9.8 m/s²)= 1.018 kgd)density of block ρ_B = m/V= (1.018 kg) / (0.107 m)³= 831.21 Pa