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Zorluk: OrtaFluids at Rest, Archimedes' Principle and Viscosity

A uniform cylindrical rod of length 20 cm20\text{ cm} floats vertically in liquid XX of density 800 kg/m3800\text{ kg/m}^3 with 15 cm15\text{ cm} of its length submerged. When transferred to liquid YY, it floats vertically with 12 cm12\text{ cm} of its length submerged. What is the density of liquid YY?

  1. A
    640 kg/m3640\text{ kg/m}^3
  2. 1000 kg/m31000\text{ kg/m}^3Cevap
  3. C
    1333 kg/m31333\text{ kg/m}^3
  4. D
    500 kg/m3500\text{ kg/m}^3

Cevap

The density of liquid YY is 1000 kg/m31000\text{ kg/m}^3.
According to the Law of Flotation, a floating body displaces its own weight of liquid. Therefore, hXρX=hYρYh_X \rho_X = h_Y \rho_Y. Substituting 15 cm×800 kg/m3=12 cm×ρY15\text{ cm} \times 800\text{ kg/m}^3 = 12\text{ cm} \times \rho_Y yields ρY=1000 kg/m3\rho_Y = 1000\text{ kg/m}^3.

Adım Adım Çözüm

1
Apply the Law of Flotation for a floating body of uniform cross-sectional area AA.
Weight of rod W=Upthrust=A⋅hsubmerged⋅ρliquid⋅gW = \text{Upthrust} = A \cdot h_{\text{submerged}} \cdot \rho_{\text{liquid}} \cdot g.
A floating body displaces a weight of fluid equal to its own total weight.
2
Equate the upthrust in liquid XX to the upthrust in liquid YY.
A⋅hX⋅ρX⋅g=A⋅hY⋅ρY⋅g  ⟹  hX⋅ρX=hY⋅ρYA \cdot h_X \cdot \rho_X \cdot g = A \cdot h_Y \cdot \rho_Y \cdot g \implies h_X \cdot \rho_X = h_Y \cdot \rho_Y.
Since the rod is identical and floating freely in both liquids, its weight WW remains unchanged.
3
Substitute the known values (hX=15 cmh_X = 15\text{ cm}, ρX=800 kg/m3\rho_X = 800\text{ kg/m}^3, hY=12 cmh_Y = 12\text{ cm}) and solve for ρY\rho_Y.
ρY=hX⋅ρXhY=15×80012=1000 kg/m3\rho_Y = \frac{h_X \cdot \rho_X}{h_Y} = \frac{15 \times 800}{12} = 1000\text{ kg/m}^3.
Rearranging the linear equation yields the density of liquid YY.

Anahtar Kavram

Law of Flotation and Hydrometer Principle
Tahmini Süre:1m 15s
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Konular

Fluids at Rest, Archimedes' Principle and Viscosity

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