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Test: Archimedes Principle - 2 - EmSAT Achieve MCQ


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Test: Archimedes Principle - 2 - Question 1

An object is put one by one in three liquids P, Q, and R having different densities. Consider the following statements and arrange the liquids in ascending order of densities.

(A) In liquid P, the object is completely submerged

(B) In liquid Q, objects float with (1/4)th part of their volume inside the liquid

(C) In liquid R, objects float with (1/2)th of their volume inside the liquid

Detailed Solution for Test: Archimedes Principle - 2 - Question 1

Concept:

Archimedes principle:

Principle of Archimedes states “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it.”

  • Thus, when a solid is fully immersed in a liquid, it loses weight which is equal to the weight of the liquid it displaces.

Apparent weight = Actual weight - buoyant force = mg - ρgV

Where m is the mass of the object, ρ is the density of the fluid

  • Hence if there is no gravity, there is no Upward Thrust.
  • The upward force applied to the object is called the buoyant force.

Here,

ρ = density of liquid

g = acceleration due to gravity

V = volume of liquid displaced by the object

Explanation:

  • When a body is immersed partially or completely in a fluid (liquid or gas), it experiences an upthrust or buoyant force that is equal to the weight of the fluid displaced by the body.
  • So the upthrust on the ball is given as,

⇒ FB = ρVg

⇒ FB ∝ ρ      -----(1)

Where ρ = density of the liquid, V = volume of liquid displaced = volume of ball, and g = gravitational acceleration

  • Buoyant Force is when a rigid object is submerged in a fluid (completely or partially), and there exits an upward force on the object that is equal to the weight of the fluid that is displaced by the object.
  • In liquid P, the object is completely submerged- The volume of the fluid displaced is equal to the volume of the object.
  • In liquid Q, objects float with (1/4)th part of their volume inside the liquid- Here, (1/4) th means 0.25.
  •  In liquid R, objects float with (1/2)th of their volume inside the liquid - Here, (1/2) means 0.50
  • As density is inversely proportional to volume, liquid Q has less volume so density will be high as compared to R and P.

So, it is proven that  P < R < Q.

Additional Information

  • Archimedes Principle is also known as the physical law of buoyancy.
  • The purity of gold in the king's crown, Archimedes determines after discovering the Archimedes principle.
  • Hydrometers, Ships, and Submarines work on the Archimedes Principle.
  • A Ship/boat floats on the basis of the Archimedes Principle.
Test: Archimedes Principle - 2 - Question 2

When the weight of the fluid displaced is less than the weight of the solid body then:

Detailed Solution for Test: Archimedes Principle - 2 - Question 2

Archimedes principle: 

  • It is used to explain the law of flotation or upward thrust experienced when immersed in a fluid. 
  • The principle of Archimedes states “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it.”
  • The flotation and sinking of an object are dependent upon the relative density with each other.
    • If the density of the object is more than the density of the liquid, the object will sink.
    • On the other side, if the density of an object is less than the liquid, then it will float over it.
    • If the density of the object and liquid is equal to each other, they are in equilibrium and float and sink both at the same time
    • Therefore, the sinking or floating of an object depends on the difference in density of objects and liquid.

Weight = m × g = ρ × g × V

  • If the upthrust force (equal to weight) exerted on the body by the fluid is greater than the weight of the body, then the body floats. While, if upthrust is lesser, then the body sinks.

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Test: Archimedes Principle - 2 - Question 3

A piece of brass (alloy of copper and zinc) weighs 12.9 g in air. When completely immersed in water weighs 11.3 g. what is the mass of copper contained in the alloy? Specific gravities of copper and zinc are 8.9 and 7.1 respectively.

Detailed Solution for Test: Archimedes Principle - 2 - Question 3

Concept:

  • The Specific gravity of a substance is the ratio of its density to the density of water.

  • A substance with a specific gravity of less than 1 will float in water. For example, an ice cube, with a relative density of about 0.91, will float. A substance with a specific gravity greater than 1 will sink.
  • Whereas the relative density of a substance is defined as the ratio of densities of any two-sample, it can be expressed as

  • Archimedes' principle states that when a body immersed in a fluid, whether fully or partially submerged, the upward buoyant force that is exerted on it, is equal to the weight of the fluid that the body displaces.
  • Buoyant force: When an object is submerged in fluid fully or partially, an upward force is exerted on the body. This force is known as the buoyant force. 

Calculation:

Given:

Mass of water displaced = 12.9 - 11.3 = 1.6 g

⇒ Volume of water displaced = mass / density = 1.6/1 = 1.6 cm3

We know that,

Relative density = density of substance / density of water

Since density of water = 1 g/cm3

⇒ Density of copper = 8.9 g/cm3

Let the mass of copper in 12.9 gram of alloy is x

Volume of water displaced by the copper  = x/8.9 cm3

Similarly,

⇒ Density of zinc = 7.1 g/cm3

Also, the mass of zinc in alloy = 12.9 - x

Volume of water displaced by zinc = (12.9- x)/7.1 cm3

Hence total water displaced by the alloy,

Solving we get,

x = 7.614 gm.

Hence the mass of copper in the alloy is 7.614 gm.

Test: Archimedes Principle - 2 - Question 4

What is the specific gravity of a marble stone, which weighs 400 N in air, and 200 N in the water? (g = 10m/s2)

Detailed Solution for Test: Archimedes Principle - 2 - Question 4

Concept:

The weight of the stone in the air is given by, ρs × V × g

Where ρs is the density of the stone and V is the volume of the stone.

When the stone is immersed in the water there will be buoyancy force acting upon the stone applied by water.

Then the weight of the stone in the water is given by Weight in air – Buoyancy force upon the stone

Buoyancy force = ρw × V × g

Where ρw is the density of water and V is that volume which is immersed in water

The specific gravity of stone

Calculation:

Given Ws-air = 400 N

then ρs × V × g = 400 N

∴ ρs × V = 40 N      ----(1)

Now when the stone is immersed in water then Ws-water = 200 N

ρs × V × g - ρw × V × g = 200 N

s - ρw) × V = 20 N      ----(2)

Now dividing eqn. (2) by eqn. (1)

∴ The specific gravity of the stone = 2

Test: Archimedes Principle - 2 - Question 5

A crane is used to lower weights into the sea (seawater density = 1025 kg/m3) for an underwater construction project. What is the percentage reduction in the tension in the rope of the crane due to a rectangular 0.4 m × 0.4 m × 3 m concrete block (density = 2300 kg/m3) when it is completely immersed in water compared to the tension in the rope when it was suspended in air

Detailed Solution for Test: Archimedes Principle - 2 - Question 5

Tension in air (T1) = mg = 

When immersed in the water then tension in water (T2) = 

% reduction in tension in rope is

Test: Archimedes Principle - 2 - Question 6

A wooden plank (sp.gr 0.5) 1 m × 1 m × 0.5 m floats is water with 1.5 kN load on it with 1 m × 1 m surface horizontal. The depth of plank lying below water surface shall be:

Detailed Solution for Test: Archimedes Principle - 2 - Question 6

Concept:

A wooden plank of W weight is floating on water as shown in figure.

The weight W and external load of 1.5 kN is acting downward and buoyancy force acts by water is in upward direction.

So by equating the forces, we will get, 1.5 + W = FB

Calculation:

Given:

Volume of plank V = 1 × 1 × 0.5 = 0.5 m3, Submerged volume Vs = 1 × 1 × h = h m3

So, from equation (1),

1.5 + (ρgV) = (ρgVs)

1.5 + (0.5 × 0.5 × 9.81) = 1 × h × 9.81

h = 0.4 m

Test: Archimedes Principle - 2 - Question 7

An object X sinks in liquids P and Q but floats in liquid R. Which of the following conclusions can be made from these observations?

A) Density of X is more than the density of P

B) Density of X is less than the density of R

C) Density of P is equal to the density of R

D) Density of P is less than the density of R

Detailed Solution for Test: Archimedes Principle - 2 - Question 7

Concept:

Archimedes principle:

  •  It states that a body, when wholly or partially immersed in liquid, experiences an upward thrust that is equal to the volume of the liquid.
  • Archimedes Principle is also known as the physical law of buoyancy.
  • When a solid is fully immersed in a liquid, it loses weight, which is equal to the weight of the liquid it displaces.

Apparent weight= actual weight−buoyant force= mg−ρgV

Where m = mass of the object and ρ = density of the fluid

 

 Buoyant force:

  • ​When a body is wholly or partially immersed in a fluid at rest, the fluid exerts pressure on the surface of the body in contact with the fluid.
  • The pressure is greater on lower surfaces of the body than on the upper surfaces as the pressure in a fluid increases with depth.
  • The resultant of all the forces is an upward force called buoyant force

Explanation:

Law of Floatation:

  • When an object is placed on liquid, an upthrust acts on it.
  • This upthrust is more than weight when the density of the object is less than weight. The object floats in this case.
  • If the density of the object is more, the object will sink as the weight will be more than upthrust.
  • When a body is immersed partially or completely in a fluid (liquid or gas), it experiences an upthrust or buoyant force that is equal to the weight of the fluid displaced by the body.
  • Considering, the question statement:
    • ρ> ρX > ρP

So, from the above relation options A, B, and D are the correct ones.

Test: Archimedes Principle - 2 - Question 8

Submarines works on which of the following principle? 

Detailed Solution for Test: Archimedes Principle - 2 - Question 8

Archimedes Principle/Law

  • Archimedes' principle states that the upward force which is exerted by an object that is either partially or fully immersed in a fluid is equal to the weight of the liquid displaced by the object. This principle helps to prove the Law of floatation.
  • The law of floatation states that the weight of a floating body is equal to the weight of the liquid displaced by it.
  • For Example- A ship or submarine in the ocean does not sink until the weight of the water it displaces is equal to its own weight

Principle of floatation

When a body displaces a weight of water equal to its own weight, it floats. This is known as the principle of floatation.

FB = weight of fluid displaced = weight of the body

Additional Information

Pascal’s Law

In a static fluid pressure at a point is equal in all directions.

Conversely, if pressure is applied to a static fluid, it will be transmitted equally in all directions.

Example: hydraulic lift, hydraulic press and hydraulic brakes.

It can also be applied to flowing fluid if the fluid is ideal.

Bernoulli’s theorem

According to Bernoulli's theorem, in a steady, incompressible, and irrotational flow the sum of pressure energy, kinetic energy, and potential energy per unit mass is always constant.

Darcy's Law

The Darcy - Weisbach equation is one of the most commonly used equations for estimating frictional pressure drops in internal flows.

Where D is the hydraulic diameter.

D = 4A/P = 4m   (A = area, P = wetted perimeter)

Test: Archimedes Principle - 2 - Question 9

Consider the following statement 

Assertion: A human floats easily by using a rubber tube.

Reason: An inflated rubber tube has low weight and large volume and increases the upthrust.

Detailed Solution for Test: Archimedes Principle - 2 - Question 9

CONCEPT:

Archimedes principle:

  •  It states that a body, when wholly or partially immersed in liquid, experiences an upward thrust which is equal to the volume of the liquid.
  • Archimedes Principle is also known as the physical law of buoyancy.
  • When a solid is fully immersed in a liquid, it loses weight, which is equal to the weight of the liquid it displaces.
  • Apparent weight= actual weight−buoyant force= mg−ρgV.

Where m = mass of the object and ρ = density of the fluid

 

Laws of Floatation: 

The three conditions required for the body to float on water are:

  1. The weight of the body is equal to the buoyant force.

  2. The weight of the body is greater than the weight of the fluid displaced

  3. The weight of the body is less than the weight of the fluid displaced.

EXPLANATION:

  • As per Archimedes principle, when wholly or partially immersed in liquid, experiences an upward thrust that is equal to the volume of the liquid.
  • A human body is slightly denser than water.
  • An inflated rubber tube has low weight. 
  • The volume of the inflated rubber tube is more. Hence, more water is displaced.
  • That increases the upthrust. It helps a person to float.
  • Hence, Both assertion and reason are correct. The reason gives the correct explanation about the assertion.

So, option 2nd is correct.

Test: Archimedes Principle - 2 - Question 10

On which of the following factors the magnitude of the buoyant force acting on a body in a given fluid depends?

Detailed Solution for Test: Archimedes Principle - 2 - Question 10

ey Points

  • The buoyant force depends on the volume of the liquid displaced. The buoyant force depends on the mass of the object, the weight of the object, and its density.
  • The buoyant force depends directly upon:
    • The volume of the fluid displaced.
    • The density of the fluid in which the body is immersed.
    • Acceleration due to gravity at the place.

Important Points

  • The three types of buoyancy are positive buoyancy, negative buoyancy, and neutral buoyancy.
    • Positive buoyancy is when the immersed object is lighter than the fluid displaced and this is the reason why the object floats.
    • Negative buoyancy is when the immersed object is denser than the fluid displaced which results in the sinking of the object.
    • Neutral buoyancy takes place when the weight of an immersed object is equal to the fluid displaced
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