We read your photo as:

Q5A. Define the following i. Hydrostatics ii. Total pressure iii. Centre pressure iv. Meta-centre v. Meta-centre height B. Define flotation and state the principle of flotation. C. Calculate the buoyant force acting on an object with a displaced volume of 2.5m³ in sea water (density =1025Kg/m³).

Physics

ScanSolve AI Answer

Answer

v

25138.125 N\text{25138.125 N}

Step-by-step solution

3 steps

Here are the solutions for Q4B, Q5A, Q5B, and Q5C.

Q4B. Calculate the density, specific weight and specific gravity of one litre of petrol of specific gravity = 0.7

Step 1: Identify the given values and constants.
Given:
Specific gravity (SGSG) of petrol = 0.7
Volume of petrol (VV) = 1 litre = 0.001 m30.001 \, \text{m}^3
Density of water (ρwater\rho_{\text{water}}) = 1000 kg/m31000 \, \text{kg/m}^3
Acceleration due to gravity (gg) = 9.81 m/s29.81 \, \text{m/s}^2

Step 2: Calculate the density of petrol.
The specific gravity is the ratio of the density of the substance to the density of water.

SG=ρpetrolρwaterSG = \frac{\rho_{\text{petrol}}}{\rho_{\text{water}}} ρpetrol=SG×ρwater\rho_{\text{petrol}} = SG \times \rho_{\text{water}} ρpetrol=0.7×1000 kg/m3\rho_{\text{petrol}} = 0.7 \times 1000 \, \text{kg/m}^3 ρpetrol=700 kg/m3\rho_{\text{petrol}} = 700 \, \text{kg/m}^3

Step 3: Calculate the specific weight of petrol.
Specific weight (γ\gamma) is the product of density and acceleration due to gravity.

γpetrol=ρpetrol×g\gamma_{\text{petrol}} = \rho_{\text{petrol}} \times g γpetrol=700 kg/m3×9.81 m/s2\gamma_{\text{petrol}} = 700 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 γpetrol=6867 N/m3\gamma_{\text{petrol}} = 6867 \, \text{N/m}^3

Step 4: State the specific gravity.
The specific gravity is given in the problem.

SGpetrol=0.7SG_{\text{petrol}} = 0.7

The density of petrol is 700 kg/m3\boxed{\text{700 kg/m}^3}.
The specific weight of petrol is 6867 N/m3\boxed{\text{6867 N/m}^3}.
The specific gravity of petrol is 0.7\boxed{\text{0.7}}.

Q5A. Define the following:

Part i

Hydrostatics

Hydrostatics is the branch of fluid mechanics that deals with the study of fluids at rest. It focuses on the pressure exerted by fluids and the forces acting on submerged or floating objects.

Part ii

Total pressure

Total pressure refers to the total force exerted by a static fluid on a submerged or partially submerged surface. It is calculated by integrating the pressure distribution over the entire area of the surface.

Part iii

Centre pressure

The centre of pressure is the point on a submerged surface where the resultant (total) hydrostatic force acts. If a single force were to replace the distributed pressure, it would act at this point.

Part iv

Meta-centre

The metacentre is a theoretical point used in the stability analysis of floating bodies. It is the point of intersection of the vertical line passing through the center of buoyancy (when the body is tilted by a small angle) and the original vertical line passing through the center of buoyancy (when the body is in its upright position).

Part v

25138.125 N\text{25138.125 N}

Meta-centre height

Metacentric height (GMGM) is the vertical distance between the metacentre (M) and the center of gravity (G) of a floating body. It is a crucial indicator of a vessel's initial stability; a positive metacentric height indicates stable equilibrium.

Q5B. Define flotation and state the principle of flotation.

Flotation is the phenomenon where an object remains suspended on the surface of a fluid, partially or wholly submerged, without sinking.

The Principle of Flotation states that for an object to float, the buoyant force acting on it must be equal to the weight of the object. This means that a floating object displaces a weight of fluid equal to its own weight.

Q5C. Calculate the buoyant force acting on an object with a displaced volume of 2.5 m32.5 \, \text{m}^3 in sea water (density = 1025 kg/m31025 \, \text{kg/m}^3).

  1. Identify the given values and constants.

    Given:
    Displaced volume (VdisplacedV_{\text{displaced}}) = 2.5 m32.5 \, \text{m}^3
    Density of sea water (ρsea water\rho_{\text{sea water}}) = 1025 kg/m31025 \, \text{kg/m}^3
    Acceleration due to gravity (gg) = 9.81 m/s29.81 \, \text{m/s}^2

  2. State the formula for buoyant force.

    According to Archimedes' Principle, the buoyant force (FBF_B) is given by:

    FB=ρfluidVdisplacedgF_B = \rho_{\text{fluid}} V_{\text{displaced}} g

  3. Substitute the values and calculate the buoyant force.

    FB=(1025 kg/m3)×(2.5 m3)×(9.81 m/s2)F_B = (1025 \, \text{kg/m}^3) \times (2.5 \, \text{m}^3) \times (9.81 \, \text{m/s}^2) FB=2562.5 kg×9.81 m/s2F_B = 2562.5 \, \text{kg} \times 9.81 \, \text{m/s}^2 FB=25138.125 NF_B = 25138.125 \, \text{N}

    The buoyant force on the object is 25138.125 N\boxed{\text{25138.125 N}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QA block of uniform cross-sectional area floats in two liquids A and B. Given densities and submerged…QCalculate the hydrostatic force on a vertical plane submerged in two layers of liquid (oil and…QState the principal of flotationQA center of pressure of a fluid or the centre of pressure of a surface to fluid pressure is the…QWhat are the basic principles of buoyancy and how do they apply to everyday examples like balloons…
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp