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Physics - Fluids (Thrust, Pressure, Archimedes' Principle, Floatation)

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Thrust is the total force acting normally (perpendicularly) on a surface. Its S.I. unit is Newton (NN) and C.G.S. unit is dyne. 1N=105 dyne1 N = 10^5 \text{ dyne}.

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Pressure is the thrust per unit area (P=FAP = \frac{F}{A}). The S.I. unit is Pascal (PaPa), where 1Pa=1Nm−21 Pa = 1 N m^{-2}.

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Pressure in fluids (liquids and gases) increases with depth (hh) and density (ρ\rho) of the fluid, acting equally in all directions at a given depth.

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Upthrust or Buoyant Force is the upward force exerted by a fluid on an object immersed in it. It depends on the volume of the submerged part of the body (VV), the density of the fluid (ρ\rho), and acceleration due to gravity (gg).

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Archimedes' Principle states that when a body is immersed partially or completely in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it.

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Relative Density (R.D.R.D.) of a substance is the ratio of its density to the density of water at 4∘C4^{\circ}C. Since it is a ratio, it has no units.

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Law of Floatation: A body floats in a liquid if the weight of the body is equal to the weight of the liquid displaced by its submerged part. This implies the apparent weight of a floating body is zero.

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Relationship for floatation: vV=ρsρl\frac{v}{V} = \frac{\rho_s}{\rho_l}, where vv is the submerged volume, VV is total volume, ρs\rho_s is density of the solid, and ρl\rho_l is density of the liquid.

📐Formulae

P=FAP = \frac{F}{A}

P=hρgP = h \rho g

FB=VρgF_B = V \rho g

R.D.=Density of substanceDensity of water at 4∘CR.D. = \frac{\text{Density of substance}}{\text{Density of water at } 4^{\circ}C}

R.D.=Weight of body in airWeight of body in air−Weight of body in waterR.D. = \frac{\text{Weight of body in air}}{\text{Weight of body in air} - \text{Weight of body in water}}

Total Pressure=Patmosphere+hρg\text{Total Pressure} = P_{atmosphere} + h \rho g

💡Examples

Problem 1:

A rectangular block of dimensions 2m×1m×0.5m2 m \times 1 m \times 0.5 m and mass 500kg500 kg is placed on the ground. Calculate the maximum pressure it can exert.

Solution:

m=500kgm = 500 kg, so Thrust F=mg=500×9.8=4900NF = mg = 500 \times 9.8 = 4900 N. Maximum pressure occurs when the contact area AA is minimum. Amin=1m×0.5m=0.5m2A_{min} = 1 m \times 0.5 m = 0.5 m^2. Pmax=FAmin=49000.5=9800PaP_{max} = \frac{F}{A_{min}} = \frac{4900}{0.5} = 9800 Pa.

Explanation:

Pressure is inversely proportional to area. To maximize pressure, the block must rest on its smallest face.

Problem 2:

A body weighs 50gf50 gf in air and 42gf42 gf when fully immersed in water. Find the volume of the body and its R.DR.D.

Solution:

Loss in weight = 50gf−42gf=8gf50 gf - 42 gf = 8 gf. Since density of water is 1gcm−31 g cm^{-3}, the upthrust of 8gf8 gf corresponds to a displaced volume of 8cm38 cm^3. Thus, Volume V=8cm3V = 8 cm^3. R.D.=Weight in airLoss in weight=508=6.25R.D. = \frac{\text{Weight in air}}{\text{Loss in weight}} = \frac{50}{8} = 6.25.

Explanation:

According to Archimedes' Principle, the loss of weight in water is equal to the weight of the water displaced. In C.G.S, the weight of water in gfgf is numerically equal to its volume in cm3cm^3.

Problem 3:

An iceberg of density 900kgm−3900 kg m^{-3} floats in sea water of density 1030kgm−31030 kg m^{-3}. What fraction of the iceberg is above the water surface?

Solution:

Let VV be total volume and vv be submerged volume. Using vV=ρsρl=9001030≈0.87\frac{v}{V} = \frac{\rho_s}{\rho_l} = \frac{900}{1030} \approx 0.87. Submerged fraction is 87%87\%. Fraction above water =1−0.87=0.13= 1 - 0.87 = 0.13 or 13%13\%.

Explanation:

The ratio of densities determines the fraction of the volume that stays submerged to balance the weight.