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Zorluk: ZorFluids at Rest, Archimedes' Principle and Viscosity

A solid cylinder of length 10 cm10\text{ cm} floats vertically at the interface of two immiscible liquids: oil of density 800 kg/m3800\text{ kg/m}^3 and water of density 1000 kg/m31000\text{ kg/m}^3. If 4 cm4\text{ cm} of its length is submerged in water while the remaining 6 cm6\text{ cm} is submerged in oil, what is the density of the cylinder?

  1. 880 kg/m3880\text{ kg/m}^3Cevap
  2. B
    920 kg/m3920\text{ kg/m}^3
  3. C
    900 kg/m3900\text{ kg/m}^3
  4. D
    400 kg/m3400\text{ kg/m}^3

Cevap

The density of the cylinder is 880 kg/m3880\text{ kg/m}^3.
According to the Law of Flotation, a floating body displaces its own weight of the fluids in which it floats. For a body submerged across two immiscible liquids, the total buoyant force is the sum of the upthrusts from both fluids: U=(ρwVw+ρoVo)gU = (\rho_w V_w + \rho_o V_o)g. Equating this to the total weight W=ρcVtotalgW = \rho_c V_{total} g yields ρc(0.10 m)=1000(0.04 m)+800(0.06 m)=88\rho_c (0.10\text{ m}) = 1000(0.04\text{ m}) + 800(0.06\text{ m}) = 88, which gives ρc=880 kg/m3\rho_c = 880\text{ kg/m}^3.

Adım Adım Çözüm

1
Set up the condition for flotation.
Weight of the cylinder = Total upthrust exerted by both liquids.
For a body floating in equilibrium, its total weight is balanced by the sum of buoyant forces from all surrounding fluids.
2
Express the weight and upthrusts in terms of density, cross-sectional area AA, length LL, and acceleration due to gravity gg.
ρcALg=(ρwAhw+ρoAho)g\rho_c A L g = (\rho_w A h_w + \rho_o A h_o) g
The weight of the cylinder is ρcVtotalg\rho_c V_{total} g and upthrust from each liquid is ρliquidVsubmergedg\rho_{liquid} V_{submerged} g.
3
Cancel common factors AA and gg from both sides.
ρcL=ρwhw+ρoho\rho_c L = \rho_w h_w + \rho_o h_o
Since the cylinder has a uniform cross-sectional area, volume ratio simplifies to length ratio.
4
Substitute the given values (L=0.10 mL = 0.10\text{ m}, hw=0.04 mh_w = 0.04\text{ m}, ho=0.06 mh_o = 0.06\text{ m}, ρw=1000 kg/m3\rho_w = 1000\text{ kg/m}^3, ρo=800 kg/m3\rho_o = 800\text{ kg/m}^3).
ρc(0.10)=1000(0.04)+800(0.06)=40+48=88 kg/m2\rho_c (0.10) = 1000(0.04) + 800(0.06) = 40 + 48 = 88\text{ kg/m}^2
Evaluating the weighted contribution of buoyancy from each fluid.
5
Solve for the density of the cylinder ρc\rho_c.
ρc=880.10=880 kg/m3\rho_c = \frac{88}{0.10} = 880\text{ kg/m}^3
Dividing both sides by the total length of 0.10 m0.10\text{ m} gives the density.

Anahtar Kavram

Archimedes' Principle for Floating Bodies in Immiscible Liquids
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Konular

Fluids at Rest, Archimedes' Principle and Viscosity

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