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All questions of Fluid Mechanics for Mechanical Engineering Exam
Which one is in a state of failure?- a)Solid
- b)Liquid
- c)Gas
- d)Fluid
Correct answer is option 'D'. Can you explain this answer?
Which one is in a state of failure?
a)
Solid
b)
Liquid
c)
Gas
d)
Fluid
|
Swara Dasgupta answered |
A fluid is a Tresca material with zero cohesion. In simple words, fluid is in a state of failure.
Viscosity of water in comparison to mercury is
- a)lower
- b)higher
- c)same
- d)higher/lower depending on temperature
- e)unpredictable
Correct answer is option 'A'. Can you explain this answer?
Viscosity of water in comparison to mercury is
a)
lower
b)
higher
c)
same
d)
higher/lower depending on temperature
e)
unpredictable
|
Rahul Chatterjee answered |
Viscosity of water in comparison to mercury is lower.
The relation between shear stress Z and velocity gradient 
of a fluid is given by 
where A and n are constants. If n = 1, what type of fluid will it be?- a)Newtonian fluid
- b)Non-Newtonian fluid
- c)Pseudoplastic
- d)Bingham plastic
Correct answer is option 'A'. Can you explain this answer?
The relation between shear stress Z and velocity gradient of a fluid is given by where A and n are constants. If n = 1, what type of fluid will it be?
of a fluid is given by where A and n are constants. If n = 1, what type of fluid will it be?a)
Newtonian fluid
b)
Non-Newtonian fluid
c)
Pseudoplastic
d)
Bingham plastic
|
Vertex Academy answered |
Explanation: When n = 1, the relation reduces to Newton’s law of viscosity: z = A * 
, where A will represent the viscosity of the fluid. The fluid following this relation will be a Newtonian fluid as it is a linear relation. The equation given is of a straight line.
, where A will represent the viscosity of the fluid. The fluid following this relation will be a Newtonian fluid as it is a linear relation. The equation given is of a straight line.Density of water is maximum at- a)0ºC
- b)0ºK
- c)4ºC
- d)100ºC
- e)20ºC
Correct answer is option 'C'. Can you explain this answer?
Density of water is maximum at
a)
0ºC
b)
0ºK
c)
4ºC
d)
100ºC
e)
20ºC
|
Mrinalini Sen answered |
3.98
Water. An especially notable irregular maximum density is that of water, which reaches a density peak at 3.98 degreeC (39.16 degree F).
In which type of matter, one won’t find a free surface?- a)Solid
- b)Liquid
- c)Gas
- d)Fluid
Correct answer is option 'C'. Can you explain this answer?
In which type of matter, one won’t find a free surface?
a)
Solid
b)
Liquid
c)
Gas
d)
Fluid
|
Anirban Khanna answered |
Solid molecules have a definite shape due to large inter-molecular forces. In liquids, molecules are free to move inside the whole mass but rarely escape from itself. Thus, liquids can form free surfaces under the effect of gravity. But, in case of gases, molecules tend to escape due to low forces of attraction. Thus, gases won’t form any free surface.
The value of the surface tension of an ideal fluid is- a)zero
- b)unity
- c)infinity
- d)more than that of a real fluid
Correct answer is option 'A'. Can you explain this answer?
The value of the surface tension of an ideal fluid is
a)
zero
b)
unity
c)
infinity
d)
more than that of a real fluid
|
Nitya Nambiar answered |
Explanation: Ideal fluids haze zero surface tension but real fluids have some finite value of surface tension.
If the surface of liquid is convex, then- a)cohesion pressure is negligible
- b)cohesion pressure is decreased
- c)cohesion pressure is increased
- d)there is no cohesion pressure
- e)none of the above
Correct answer is option 'C'. Can you explain this answer?
If the surface of liquid is convex, then
a)
cohesion pressure is negligible
b)
cohesion pressure is decreased
c)
cohesion pressure is increased
d)
there is no cohesion pressure
e)
none of the above
|
Zoya Sharma answered |
Correct Answer :- c
Explanation : If the surface of the liquid is convex, then the cohesion pressure will be increased. A convex meniscus occurs when the particles in the liquid have stronger attraction to each other than the material of the container.
Which of the following meters is not associated with viscosity ?- a)Red wood
- b)Say bolt
- c)Engler
- d)Orsat
- e)none of the above
Correct answer is 'D'. Can you explain this answer?
Which of the following meters is not associated with viscosity ?
a)
Red wood
b)
Say bolt
c)
Engler
d)
Orsat
e)
none of the above
|
|
Mrinalini Sen answered |
An Orsat gas analyser is a piece of laboratory equipment used to analyse a gas sample (typically fossil fuel flue gas) for its oxygen, carbon monoxide and carbon dioxide content.
Can you explain the answer of this question below:Fluid is a substance which offers no resistance to change of- A:pressure
- B:flow
- C:shape
- D:volume
- E:temperature
The answer is c.
Fluid is a substance which offers no resistance to change of
A:
pressure
B:
flow
C:
shape
D:
volume
E:
temperature
|
Baishali Bajaj answered |
Because fluids (mostly liquids and gases) do not have a definite shape, and they assume the shape of the containing vessel.
The value of the viscosity of an ideal fluid is
- a) zero
- b)unity
- c)infinity
- d)more than that of a real fluid
Correct answer is option 'A'. Can you explain this answer?
The value of the viscosity of an ideal fluid is
a)
zero
b)
unity
c)
infinity
d)
more than that of a real fluid
|
|
Manasa Sen answered |
Correct option is A.
A fluid that has no resistance to shear stress between its layers is known as an ideal fluid. Hence ideal fluid has zero viscosity.
A fluid that has no resistance to shear stress between its layers is known as an ideal fluid. Hence ideal fluid has zero viscosity.
The pressure at a point in a fluid will not be same in all the directions when the fluid is- a)moving
- b)viscous
- c)viscous and static
- d)inviscous and moving
- e)viscous and moving.
Correct answer is option 'E'. Can you explain this answer?
The pressure at a point in a fluid will not be same in all the directions when the fluid is
a)
moving
b)
viscous
c)
viscous and static
d)
inviscous and moving
e)
viscous and moving.
|
Shreus Shinde answered |
The Pascal law states that the liquid at rest applies pressure at a point is the same in all directions. This means that the pressure is there in spite of the direction. And it is present in the same direction and is having the same magnitude.
For very great pressures, viscosity of most gases and liquids- a)remains same
- b)increases
- c)decreases
- d)shows erratic behaviour
- e)none of the above
Correct answer is option 'D'. Can you explain this answer?
For very great pressures, viscosity of most gases and liquids
a)
remains same
b)
increases
c)
decreases
d)
shows erratic behaviour
e)
none of the above
|
|
Anirban Khanna answered |
For ideal gases, viscosity depends only on temperature. For real gases, that's still a very good approximation. "In most cases, a fluid's viscosity increases with increasing pressure. Compared to the temperature influence, liquids are influenced very little by the applied pressure.
An ideal flow of any fluid must fulfill the following- a)Newton's law of motion
- b)Newton's law of viscosity
- c)Pascal' law
- d)Continuity equation
- e)Boundary layer theory
Correct answer is option 'D'. Can you explain this answer?
An ideal flow of any fluid must fulfill the following
a)
Newton's law of motion
b)
Newton's law of viscosity
c)
Pascal' law
d)
Continuity equation
e)
Boundary layer theory
|
Om Desai answered |
An ideal flow of any fluid must fulfill the continuity equation i.e Av=constant
Dimensions of surface tension are- a)M1L0T–2
- b)M1L0T–1
- c)M1L1T–2
- d)M1L2T–2
- e)M1L0T1
Correct answer is option 'A'. Can you explain this answer?
Dimensions of surface tension are
a)
M1L0T–2
b)
M1L0T–1
c)
M1L1T–2
d)
M1L2T–2
e)
M1L0T1
|
|
Swara Dasgupta answered |
Surface Tension is defined as force per unit length. The formula is:
Surface Tension= Force/Length
Dimensional Formula of Force = M^1 L^1 T^-2
Divide it by Length L^1
So, Dimensional Formula of Surface Tension = M^1 L^0 T^-2
SI unit of Surface Tension is Nm-1
The specific weight of water is 1000 kg/m3- a)at normal pressure of 760 mm
- b)at 4ºC temperature
- c)at mean sea level
- d)all the above
- e)none of the above
Correct answer is option 'D'. Can you explain this answer?
The specific weight of water is 1000 kg/m3
a)
at normal pressure of 760 mm
b)
at 4ºC temperature
c)
at mean sea level
d)
all the above
e)
none of the above
|
|
Mrinalini Sen answered |
The specific weight (also known as the unit weight) is the weight per unit volume of a material. The symbol of specific weight is γ (the Greek letter Gamma). A commonly used value is the specific weight of water on Earth at 4 deg C which is 9.807 kN/m3 or 62.43 lbf/ft3.
Shear stress in static fluid is- a)always zero
- b)always maximum
- c)between zero to maximum
- d)unpredictable
Correct answer is option 'A'. Can you explain this answer?
Shear stress in static fluid is
a)
always zero
b)
always maximum
c)
between zero to maximum
d)
unpredictable
|
Raghav Mukherjee answered |
When a fluid is at rest neither shear forces nor shear stresses exist in it.
Mercury does not wet glass. This is due to property of liquid known as
- a)surface tension
- b)cohesion
- c)adhesion
- d)viscosity
- e)compressibility
Correct answer is option 'C'. Can you explain this answer?
Mercury does not wet glass. This is due to property of liquid known as
a)
surface tension
b)
cohesion
c)
adhesion
d)
viscosity
e)
compressibility
|
Shivam Sharma answered |
Formation of a Meniscus. When liquid water is confined in a tube, its surface (meniscus) has a concave shape because water wets the surface and creeps up the side. Mercury does not wet glass - the cohesive forces within the drops are stronger than the adhesive forces between the drops and glass.
For manometer, a better liquid combination is one having- a)higher surface tension
- b)lower surface tension
- c)surface tension is no criterion
- d)high density and viscosity
- e)low density and viscosity.
Correct answer is option 'A'. Can you explain this answer?
For manometer, a better liquid combination is one having
a)
higher surface tension
b)
lower surface tension
c)
surface tension is no criterion
d)
high density and viscosity
e)
low density and viscosity.
|
|
Shivam Sharma answered |
manometer liquid has high surface tension then the liquid surface in a tube will be in a shape of the dome it wouldn't be flat so we can't measure a correct reading that's why liquid should have low surface tension.
An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate is- a)0.651 × 10–3
- b)0.651
- c)6.51
- d)0.651 × 103
Correct answer is option 'B'. Can you explain this answer?
An incompressible fluid (kinematic viscosity, 7.4 × 10–7 m2/s, specific gravity, 8.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate is
a)
0.651 × 10–3
b)
0.651
c)
6.51
d)
0.651 × 103
|
|
Mrinalini Sen answered |
If a person studies about a fluid which is at rest, what will you call his domain of study?- a)Fluid Mechanics
- b)Fluid Statics
- c)Fluid Kinematics
- d)Fluid Dynamics
Correct answer is option 'B'. Can you explain this answer?
If a person studies about a fluid which is at rest, what will you call his domain of study?
a)
Fluid Mechanics
b)
Fluid Statics
c)
Fluid Kinematics
d)
Fluid Dynamics
|
|
Anand Kumar answered |
Explanation: Fluid Mechanics deals with the study of fluid at rest or in motion with or without the consideration of forces, Fluid Statics is the study of fluid at rest, Fluid Kinematics is the study of fluid in motion without consideration of forces and Fluid Dynamics is the study of fluid in motion considering the application forces.
The units of viscosity are- a)metres2 per sec
- b)kg sec/metre2
- c)newton-sec per metre2
- d)newton-sec2 per metre
- e)none of the above
Correct answer is option 'C'. Can you explain this answer?
The units of viscosity are
a)
metres2 per sec
b)
kg sec/metre2
c)
newton-sec per metre2
d)
newton-sec2 per metre
e)
none of the above
|
|
Mrinalini Sen answered |
Physical unit of dynamic viscosity in SI units, the poiseuille (Pl), and cgs units, the poise (P), are named after Jean Lonard Marie Poiseuille. The poiseuille, which is rarely used, is equivalent to the pascal second (Pa. s), or (N. s)/m2, or kg/(m.
The pressure inside a soap bubble of 50 mm diameter is 25 N/m2 above the atmospheric pressure. What is the surface tension is soap film- a)0.156 N/m
- b)0.312 N/m
- c)0.624 N/m
- d)0.948 N/m
Correct answer is option 'A'. Can you explain this answer?
The pressure inside a soap bubble of 50 mm diameter is 25 N/m2 above the atmospheric pressure. What is the surface tension is soap film
a)
0.156 N/m
b)
0.312 N/m
c)
0.624 N/m
d)
0.948 N/m
|
Mrinalini Sharma answered |
Given data:
Diameter of soap bubble, d = 50 mm
Pressure inside the bubble, P = 25 N/m²
Surface tension of soap film, T = ?
We know that the pressure inside a soap bubble is given by the Laplace’s law:
P = 2T/r
where r is the radius of the soap bubble.
Calculation:
Given, diameter of bubble, d = 50 mm
Radius of bubble, r = d/2 = 25 mm = 0.025 m
Pressure inside bubble, P = 25 N/m²
Substituting the values in Laplace’s law, we get:
25 = 2T/0.025
T = 0.025 x 25/2 = 0.3125 N/m
Therefore, the surface tension of soap film is 0.156 N/m.
Hence, the correct option is (a) 0.156 N/m.
Diameter of soap bubble, d = 50 mm
Pressure inside the bubble, P = 25 N/m²
Surface tension of soap film, T = ?
We know that the pressure inside a soap bubble is given by the Laplace’s law:
P = 2T/r
where r is the radius of the soap bubble.
Calculation:
Given, diameter of bubble, d = 50 mm
Radius of bubble, r = d/2 = 25 mm = 0.025 m
Pressure inside bubble, P = 25 N/m²
Substituting the values in Laplace’s law, we get:
25 = 2T/0.025
T = 0.025 x 25/2 = 0.3125 N/m
Therefore, the surface tension of soap film is 0.156 N/m.
Hence, the correct option is (a) 0.156 N/m.
Which of the following is a shear-thinnning fluid?- a)Bingham plastic
- b)Rheopectic
- c)Dilatant
- d)Pseudoplastic
Correct answer is option 'D'. Can you explain this answer?
Which of the following is a shear-thinnning fluid?
a)
Bingham plastic
b)
Rheopectic
c)
Dilatant
d)
Pseudoplastic
|
|
Anuj Chakraborty answered |
Explanation: Shear-thinning fluids are those which gets strained easily at high values of shear stresses. The relation between shear stress Z and velocity gradient 
of a shear-thinning fluid is given by 
, where A and n are constants and n < 1. This relation is followed by Pseudoplastics.
of a shear-thinning fluid is given by , where A and n are constants and n < 1. This relation is followed by Pseudoplastics.What will be the dimension of the flow consistency index for a fluid with a flow behaviour index of -1? - a)N/m2 s2
- b)N/m2 s
- c)N/ms
- d)N/ms2
Correct answer is option 'B'. Can you explain this answer?
What will be the dimension of the flow consistency index for a fluid with a flow behaviour index of -1?
a)
N/m2 s2
b)
N/m2 s
c)
N/ms
d)
N/ms2
|
|
Partho Jain answered |
Explanation: The relation between shear stress Z and velocity gradient 
of a fluid is given by 
where A is the flow consistency index and n is the flow behaviour index. If n = -1, A = Z * Unit of Z is N/m2 and is s-1. Thus, the unit of A will be N/m2 s.
of a fluid is given by where A is the flow consistency index and n is the flow behaviour index. If n = -1, A = Z * Unit of Z is N/m2 and is s-1. Thus, the unit of A will be N/m2 s.An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm diameter. The flow is critical at a velocity of about- a)1.5 m/s
- b)2.2 m/s
- c)2.5 m/s
- d)3 m/s
Correct answer is option 'C'. Can you explain this answer?
An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm diameter. The flow is critical at a velocity of about
a)
1.5 m/s
b)
2.2 m/s
c)
2.5 m/s
d)
3 m/s
|
Avantika Sen answered |
Critical Velocity of Flow in Pipes
The critical velocity of flow in pipes is the velocity at which the flow changes from a laminar flow to a turbulent flow. This phenomenon is known as the transition from laminar to turbulent flow.
Formula
The critical velocity of flow in pipes can be calculated using the following formula:
Vc = (Re * μ) / (ρ * D)
where
Vc = critical velocity of flow (m/s)
Re = Reynolds number
μ = kinematic viscosity of the fluid (stokes)
ρ = density of the fluid (kg/m3)
D = diameter of the pipe (m)
Solution
Given,
μ = 0.5 stokes
D = 4 cm = 0.04 m
To find the critical velocity, we need to calculate the Reynolds number (Re) first.
Re = (ρ * V * D) / μ
where
V = velocity of flow (m/s)
Assuming the fluid to be oil, with a density of 900 kg/m3, we can calculate the critical velocity as follows:
For option 'A':
Vc = (Re * μ) / (ρ * D)
Assuming V = 1.5 m/s
Re = (900 * 1.5 * 0.04) / 0.5 = 5,400
Vc = (5,400 * 0.5) / (900 * 0.04) = 1.5 m/s
For option 'B':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.2 m/s
Re = (900 * 2.2 * 0.04) / 0.5 = 17,820
Vc = (17,820 * 0.5) / (900 * 0.04) = 2.49 m/s
For option 'C':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.5 m/s
Re = (900 * 2.5 * 0.04) / 0.5 = 22,275
Vc = (22,275 * 0.5) / (900 * 0.04) = 2.77 m/s
For option 'D':
Vc = (Re * μ) / (ρ * D)
Assuming V = 3 m/s
Re = (900 * 3 * 0.04) / 0.5 = 27,000
Vc = (27,000 * 0.5) / (900 * 0.04) = 3.33 m/s
Thus, we can see that the critical velocity of flow in the given pipe is about 2.5 m/s, which is option 'C'.
The critical velocity of flow in pipes is the velocity at which the flow changes from a laminar flow to a turbulent flow. This phenomenon is known as the transition from laminar to turbulent flow.
Formula
The critical velocity of flow in pipes can be calculated using the following formula:
Vc = (Re * μ) / (ρ * D)
where
Vc = critical velocity of flow (m/s)
Re = Reynolds number
μ = kinematic viscosity of the fluid (stokes)
ρ = density of the fluid (kg/m3)
D = diameter of the pipe (m)
Solution
Given,
μ = 0.5 stokes
D = 4 cm = 0.04 m
To find the critical velocity, we need to calculate the Reynolds number (Re) first.
Re = (ρ * V * D) / μ
where
V = velocity of flow (m/s)
Assuming the fluid to be oil, with a density of 900 kg/m3, we can calculate the critical velocity as follows:
For option 'A':
Vc = (Re * μ) / (ρ * D)
Assuming V = 1.5 m/s
Re = (900 * 1.5 * 0.04) / 0.5 = 5,400
Vc = (5,400 * 0.5) / (900 * 0.04) = 1.5 m/s
For option 'B':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.2 m/s
Re = (900 * 2.2 * 0.04) / 0.5 = 17,820
Vc = (17,820 * 0.5) / (900 * 0.04) = 2.49 m/s
For option 'C':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.5 m/s
Re = (900 * 2.5 * 0.04) / 0.5 = 22,275
Vc = (22,275 * 0.5) / (900 * 0.04) = 2.77 m/s
For option 'D':
Vc = (Re * μ) / (ρ * D)
Assuming V = 3 m/s
Re = (900 * 3 * 0.04) / 0.5 = 27,000
Vc = (27,000 * 0.5) / (900 * 0.04) = 3.33 m/s
Thus, we can see that the critical velocity of flow in the given pipe is about 2.5 m/s, which is option 'C'.
A fluid in equilibrium can't sustain- a)tensile stress
- b)compressive stress
- c)shear stress
- d)bending stress
- e)all of the above
Correct answer is option 'C'. Can you explain this answer?
A fluid in equilibrium can't sustain
a)
tensile stress
b)
compressive stress
c)
shear stress
d)
bending stress
e)
all of the above
|
|
Shivam Sharma answered |
Fluid cannot sustain a shear force in equilibrium Fluids cannot sustain shear stress but fluids in motion offer shear resistance to flow which gives rise to viscosity.
Fluids cannot sustain shear stress means that when a force and thereby stress is applied on the fluid, the fluid's orientation changes. (as a consequence of stress being proportional to strain rate.)
Match List I with List II and select the correct answer.
- a)D
- b)C
- c)B
- d)A
Correct answer is option 'C'. Can you explain this answer?
Match List I with List II and select the correct answer.
a)
D
b)
C
c)
B
d)
A
|
Debolina Menon answered |
Correct Answer :- c
Explanation : a) A fluid that has no resistance to shear stress is known as an ideal or inviscid fluid. Zero viscosity is observed only at very low temperatures in superfluids. Otherwise, the second law of thermodynamics requires all fluids to have positive viscosity; such fluids are technically said to be viscous or viscid.
A Newtonian fluid is one whose viscosity is not affected by shear rate: all else being equal, flow speeds or shear rates do not change the viscosity.
The liquid which wets the wall of tube rises in the tube and the liquid which does not wet the wall of the tube descends in the tube. For example, when a glass capillary tube is dipped in water, water rises in the tube and the shape of water meniscus in concave, similarly when a glass capillary tube is dipped into mercury; mercury descends in the tube and the shape of mercury meniscus is convex.
A balloon lifting in air follows the following principle- a)law of gravitation
- b)Archimedes principle
- c)principle of buoyance
- d)all of the above
- e)continuity equation
Correct answer is option 'D'. Can you explain this answer?
A balloon lifting in air follows the following principle
a)
law of gravitation
b)
Archimedes principle
c)
principle of buoyance
d)
all of the above
e)
continuity equation
|
|
Divya Mehta answered |
Balloon lifting in air - Principle involved
Introduction:
Balloons are a popular mode of transportation during festivals and events. They are lightweight and can be filled with either helium or hot air to lift off the ground. The principle involved in the lifting of a balloon in the air is an important concept in physics.
Principles Involved:
The following principles are involved in the lifting of a balloon in the air:
1. Law of Gravitation: The law of gravitation states that any two objects in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The earth attracts the balloon downwards with a force called the weight of the balloon.
2. Archimedes Principle: Archimedes principle states that when an object is immersed in a fluid, it experiences a buoyant force that is equal to the weight of the fluid displaced by the object. When the balloon is filled with helium or hot air and released, it rises upwards due to the buoyant force acting on it.
3. Principle of Buoyancy: The principle of buoyancy states that a body immersed in a fluid experiences an upward force called the buoyant force that is equal to the weight of the fluid displaced by the body. The balloon filled with helium or hot air has a lower density than the surrounding air and hence experiences an upward buoyant force that lifts it up into the air.
Conclusion:
In conclusion, the lifting of a balloon in the air involves the principles of gravitation, Archimedes principle, and the principle of buoyancy. These principles are important concepts in physics and help us understand how objects move and behave in different fluids.
Introduction:
Balloons are a popular mode of transportation during festivals and events. They are lightweight and can be filled with either helium or hot air to lift off the ground. The principle involved in the lifting of a balloon in the air is an important concept in physics.
Principles Involved:
The following principles are involved in the lifting of a balloon in the air:
1. Law of Gravitation: The law of gravitation states that any two objects in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The earth attracts the balloon downwards with a force called the weight of the balloon.
2. Archimedes Principle: Archimedes principle states that when an object is immersed in a fluid, it experiences a buoyant force that is equal to the weight of the fluid displaced by the object. When the balloon is filled with helium or hot air and released, it rises upwards due to the buoyant force acting on it.
3. Principle of Buoyancy: The principle of buoyancy states that a body immersed in a fluid experiences an upward force called the buoyant force that is equal to the weight of the fluid displaced by the body. The balloon filled with helium or hot air has a lower density than the surrounding air and hence experiences an upward buoyant force that lifts it up into the air.
Conclusion:
In conclusion, the lifting of a balloon in the air involves the principles of gravitation, Archimedes principle, and the principle of buoyancy. These principles are important concepts in physics and help us understand how objects move and behave in different fluids.
When a block of ice floating on water in a container melts, the level of water in the container- a)rises
- b)first falls and then rises
- c)remains the same
- d)falls
Correct answer is option 'C'. Can you explain this answer?
When a block of ice floating on water in a container melts, the level of water in the container
a)
rises
b)
first falls and then rises
c)
remains the same
d)
falls
|
Neha Choudhury answered |
The water level remains the same when the ice cube melts. A floating object displaces an amount of water equal to its own weight. Since water expands when it freezes, one ounce of frozen water has a larger volume than one ounce of liquid water.
Units of surface tension are- a)energy/unit area
- b)distance
- c)Newton per metre
- d)it has no units
- e)none of the above
Correct answer is option 'C'. Can you explain this answer?
Units of surface tension are
a)
energy/unit area
b)
distance
c)
Newton per metre
d)
it has no units
e)
none of the above
|
Aditya Deshmukh answered |
Surface tension is measured in force per unit length. The SI unit is Newton per metre but the CGS unit of dyne per cm is also used.
The density of water is 1000 kg/m3 at- a)0 ºC
- b)0 ºK
- c)4 ºC
- d)20 ºC
- e)all temperature
Correct answer is option 'C'. Can you explain this answer?
The density of water is 1000 kg/m3 at
a)
0 ºC
b)
0 ºK
c)
4 ºC
d)
20 ºC
e)
all temperature
|
|
Baishali Bajaj answered |
The SI unit of density is kg/m3. Water of 4 DEGC is the reference ρ = 1000 kg/m3 = 1 kg/dm3 = 1 kg/l or 1 g/cm3 = 1 g/ml.
Units of mass density are- a)kg/km
- b)kg/m3
- c)
- d)
- e)
Correct answer is option 'B'. Can you explain this answer?
Units of mass density are
a)
kg/km
b)
kg/m3
c)
d)
e)
|
|
Baishali Bajaj answered |
The SI units of mass density are kg/m3, but there are several other common units. One of the most commonly used units of mass density is gram per cubic centimeter, or g/cc. This is because pure water has a mass density of 1 g/cc.
The stress-strain relation of the newtoneon fluid is- a)linear
- b)parabolic
- c)phyperbolic
- d)inverse type
- e)none of the above
Correct answer is option 'A'. Can you explain this answer?
The stress-strain relation of the newtoneon fluid is
a)
linear
b)
parabolic
c)
phyperbolic
d)
inverse type
e)
none of the above
|
Bijoy Kapoor answered |
In continuum mechanics, a Newtonian fluid is a fluid in which the viscous stresses arising from its flow, at every point, are linearly proportional to the local strain rate—the rate of change of its deformation over time.
There is no geometrical distinction between the streamline, pathline and streak line in case of- a)steady flow
- b)uniform flow
- c)laminar flow
- d)irrotational flow
Correct answer is option 'A'. Can you explain this answer?
There is no geometrical distinction between the streamline, pathline and streak line in case of
a)
steady flow
b)
uniform flow
c)
laminar flow
d)
irrotational flow
|
|
Neha Choudhury answered |
In a steady flow the streamline, pathline and streakline all coincide. In an unsteady flow they can be different. Streamlines are easily generated mathematically while pathline and streaklines are obtained through experiments. The following animation illustrates the differences between a streakline and a pathline.
Shear velocity is- a)a non-dimensional quantity
- b)a convenient fictious quantity
- c)the velocity of fluid at the edge of laminar sublayer
- d)the velocity of fluid at the edge of roughness
Correct answer is option 'B'. Can you explain this answer?
Shear velocity is
a)
a non-dimensional quantity
b)
a convenient fictious quantity
c)
the velocity of fluid at the edge of laminar sublayer
d)
the velocity of fluid at the edge of roughness
|
|
Rahul Chatterjee answered |
Shear Velocity, also called friction velocity, is a form by which a shear stress may be re-written in units of velocity. It is useful as a method in fluid mechanics to compare true velocities, such as the velocity of a flow in a stream, to a velocity that relates shear between layers of flow.
When subjected to shear force, a fluid :- a)deforms continuously no matter how small the shear stress may be
- b)deforms continuously only for large shear forces
- c)undergoes static deformation
- d)deforms continuously only for small shear stresses
Correct answer is option 'A'. Can you explain this answer?
When subjected to shear force, a fluid :
a)
deforms continuously no matter how small the shear stress may be
b)
deforms continuously only for large shear forces
c)
undergoes static deformation
d)
deforms continuously only for small shear stresses
|
|
Arjun Menon answered |
A fluid is defined as a substance that deforms continuously whilst acted upon by any forcetangential to the area on which it acts. Such a force is termedashear force, and theratio of the shear force to the area on which it acts is known astheshear stress
A small shear force is applied on an element and then removed. If the element regains it’s original position, what kind of an element can it be?- a)Solid
- b)Liquid
- c)Fluid
- d)Gaseous
Correct answer is option 'A'. Can you explain this answer?
A small shear force is applied on an element and then removed. If the element regains it’s original position, what kind of an element can it be?
a)
Solid
b)
Liquid
c)
Fluid
d)
Gaseous
|
|
Avinash Mehta answered |
Fluids (liquids and gases) cannot resist even a small shear force and gets permanently deformed. Hence, the element must be a solid element.
The relation between shear stress Z and velocity gradient of a fluid is given by 
where A and n are constants. The graphs are drawn for three values of n. Which one will be the correct relationship between n1, n2 and n3?
- a)n1 > n2 > n3
- b)n1 < n2 < n3
- c)n1 > n3 > n2
- d)n1 < n3 < n2
Correct answer is option 'B'. Can you explain this answer?
The relation between shear stress Z and velocity gradient of a fluid is given by where A and n are constants. The graphs are drawn for three values of n. Which one will be the correct relationship between n1, n2 and n3?
of a fluid is given by where A and n are constants. The graphs are drawn for three values of n. Which one will be the correct relationship between n1, n2 and n3?a)
n1 > n2 > n3
b)
n1 < n2 < n3
c)
n1 > n3 > n2
d)
n1 < n3 < n2
|
|
Dipanjan Ghosh answered |
Explanation: The graph corresponding to n = n1 represents Pseudoplastics, for which the rate of change of the shear stress decreases with the increase in the value of velocity gradient. The graph corresponding to n = n2 represents Newtonian fluids, for which shear stress changes linearly with the change in velocity gradient. The graph corresponding to n = n3 represents Dilatents, for which the rate of change of the shear stress increases with the increase in the value of velocity gradient.
Oil in a hydraulic cylinder is compressed from an initial volume 2 m3 to 1.96 m3. If the pressure of oil in the cylinder changes from 40 MPa to 80 MPa during compression, the bulk modulus of elasticity of oil is _____- a)1000 MPa
- b)2000 MPa
- c)4000 MPa
- d)8000 MPa
Correct answer is option 'B'. Can you explain this answer?
Oil in a hydraulic cylinder is compressed from an initial volume 2 m3 to 1.96 m3. If the pressure of oil in the cylinder changes from 40 MPa to 80 MPa during compression, the bulk modulus of elasticity of oil is _____
a)
1000 MPa
b)
2000 MPa
c)
4000 MPa
d)
8000 MPa
|
Bhargavi Kulkarni answered |
Understanding Bulk Modulus of Elasticity
The bulk modulus of elasticity (K) is a measure of a material's resistance to uniform compression. It is defined by the formula:
K = -ΔP / (ΔV/V₀)
Where:
- ΔP = Change in pressure
- ΔV = Change in volume
- V₀ = Initial volume
Given Data
- Initial volume (V₀) = 2 m³
- Final volume (V) = 1.96 m³
- Initial pressure (P₀) = 40 MPa
- Final pressure (P) = 80 MPa
Calculating the Change in Pressure
- ΔP = P - P₀ = 80 MPa - 40 MPa = 40 MPa
Calculating the Change in Volume
- ΔV = V - V₀ = 1.96 m³ - 2 m³ = -0.04 m³ (a decrease in volume)
Calculating the Initial Volume Ratio
- V₀ = 2 m³
Plugging Values into the Bulk Modulus Formula
Now we can calculate the bulk modulus:
K = -ΔP / (ΔV/V₀)
K = -40 MPa / (-0.04 m³ / 2 m³)
K = -40 MPa / (-0.02)
K = 2000 MPa
Conclusion
The calculated bulk modulus of elasticity of the oil is 2000 MPa, which confirms that the correct answer is option 'B'.
The bulk modulus of elasticity (K) is a measure of a material's resistance to uniform compression. It is defined by the formula:
K = -ΔP / (ΔV/V₀)
Where:
- ΔP = Change in pressure
- ΔV = Change in volume
- V₀ = Initial volume
Given Data
- Initial volume (V₀) = 2 m³
- Final volume (V) = 1.96 m³
- Initial pressure (P₀) = 40 MPa
- Final pressure (P) = 80 MPa
Calculating the Change in Pressure
- ΔP = P - P₀ = 80 MPa - 40 MPa = 40 MPa
Calculating the Change in Volume
- ΔV = V - V₀ = 1.96 m³ - 2 m³ = -0.04 m³ (a decrease in volume)
Calculating the Initial Volume Ratio
- V₀ = 2 m³
Plugging Values into the Bulk Modulus Formula
Now we can calculate the bulk modulus:
K = -ΔP / (ΔV/V₀)
K = -40 MPa / (-0.04 m³ / 2 m³)
K = -40 MPa / (-0.02)
K = 2000 MPa
Conclusion
The calculated bulk modulus of elasticity of the oil is 2000 MPa, which confirms that the correct answer is option 'B'.
Surface tension- a)acts in the plane of the interface normal to any line in the surface
- b)is also known as capillarity
- c)is a function of the curvature of the interface
- d)decreases with fall in temperature
- e)has no units
Correct answer is option 'A'. Can you explain this answer?
Surface tension
a)
acts in the plane of the interface normal to any line in the surface
b)
is also known as capillarity
c)
is a function of the curvature of the interface
d)
decreases with fall in temperature
e)
has no units
|
|
Shivam Sharma answered |
Surface tension, property of a liquid surface displayed by its acting as if it were a stretched elastic membrane. This phenomenon can be observed in the nearly spherical shape of small drops of liquids and of soap bubbles. Because of this property, certain insects can stand on the surface of water. A razor blade also can be supported by the surface tension of water. The razor blade is not floating: if pushed through the surface, it sinks through the water.
Capillary action is due to the- a)surface tension
- b)cohesion of the liquid
- c)adhesion of the liquid molecules and the molecules on the surface of a solid
- d)all of the above
- e)none of the above
Correct answer is option 'D'. Can you explain this answer?
Capillary action is due to the
a)
surface tension
b)
cohesion of the liquid
c)
adhesion of the liquid molecules and the molecules on the surface of a solid
d)
all of the above
e)
none of the above
|
Pathri Yaswanth Srinivas answered |
Capillarity defined as phenomenon of rise or fall of liquid in a small tube relative adjacent general level
capillary rise= 4×surface tension(x)×cos ¥/(w×d)
cos ¥ relates to the adhesive or cohesive of liquid molecules
w= specific weight
d= diameter of pipe
capillary rise= 4×surface tension(x)×cos ¥/(w×d)
cos ¥ relates to the adhesive or cohesive of liquid molecules
w= specific weight
d= diameter of pipe
In the Navier-Stokes equations, the forces considered are- a)pressure, viscous and turbulence
- b)gravity, pressure and viscous
- c)gravity, pressure and turbulence
- d)pressure, gravity, turbulence and viscous
Correct answer is option 'D'. Can you explain this answer?
In the Navier-Stokes equations, the forces considered are
a)
pressure, viscous and turbulence
b)
gravity, pressure and viscous
c)
gravity, pressure and turbulence
d)
pressure, gravity, turbulence and viscous
|
|
Meera Bose answered |
**Answer:**
The Navier-Stokes equations are a set of partial differential equations that describe the motion of fluid substances. They are derived from the fundamental principles of conservation of mass, momentum, and energy. In the Navier-Stokes equations, the forces considered are gravity, pressure, and viscous forces. Turbulence is not explicitly included as a force, but it is accounted for in the form of turbulence models.
**Gravity:**
Gravity is one of the forces considered in the Navier-Stokes equations. It is responsible for the buoyancy effect in fluid flow and contributes to the overall flow behavior, especially in vertical or inclined flows.
**Pressure:**
Pressure is another force that is considered in the Navier-Stokes equations. It plays a significant role in determining the fluid flow behavior. Pressure forces arise due to the variation in fluid pressure across the flow domain. These pressure forces can accelerate or decelerate the fluid particles, leading to changes in velocity and flow patterns.
**Viscous Forces:**
Viscous forces are also included in the Navier-Stokes equations. Viscosity is a property of fluids that quantifies their resistance to deformation under shear stress. Viscous forces arise due to the internal friction within the fluid and act to dissipate mechanical energy. These forces are responsible for the transfer of momentum between fluid layers, leading to the development of velocity gradients and the damping of flow disturbances.
**Turbulence:**
Turbulence is not explicitly included as a force in the Navier-Stokes equations. However, it is an important aspect of fluid dynamics that affects the flow behavior. Turbulence is characterized by chaotic and irregular fluctuations in velocity and pressure. It arises due to instabilities in the flow, and it can significantly impact the transport of momentum, heat, and mass. In practice, turbulence is accounted for by using turbulence models, which provide closure to the set of equations.
In summary, the forces considered in the Navier-Stokes equations are gravity, pressure, and viscous forces. Turbulence is not explicitly included as a force but is accounted for through the use of turbulence models.
The Navier-Stokes equations are a set of partial differential equations that describe the motion of fluid substances. They are derived from the fundamental principles of conservation of mass, momentum, and energy. In the Navier-Stokes equations, the forces considered are gravity, pressure, and viscous forces. Turbulence is not explicitly included as a force, but it is accounted for in the form of turbulence models.
**Gravity:**
Gravity is one of the forces considered in the Navier-Stokes equations. It is responsible for the buoyancy effect in fluid flow and contributes to the overall flow behavior, especially in vertical or inclined flows.
**Pressure:**
Pressure is another force that is considered in the Navier-Stokes equations. It plays a significant role in determining the fluid flow behavior. Pressure forces arise due to the variation in fluid pressure across the flow domain. These pressure forces can accelerate or decelerate the fluid particles, leading to changes in velocity and flow patterns.
**Viscous Forces:**
Viscous forces are also included in the Navier-Stokes equations. Viscosity is a property of fluids that quantifies their resistance to deformation under shear stress. Viscous forces arise due to the internal friction within the fluid and act to dissipate mechanical energy. These forces are responsible for the transfer of momentum between fluid layers, leading to the development of velocity gradients and the damping of flow disturbances.
**Turbulence:**
Turbulence is not explicitly included as a force in the Navier-Stokes equations. However, it is an important aspect of fluid dynamics that affects the flow behavior. Turbulence is characterized by chaotic and irregular fluctuations in velocity and pressure. It arises due to instabilities in the flow, and it can significantly impact the transport of momentum, heat, and mass. In practice, turbulence is accounted for by using turbulence models, which provide closure to the set of equations.
In summary, the forces considered in the Navier-Stokes equations are gravity, pressure, and viscous forces. Turbulence is not explicitly included as a force but is accounted for through the use of turbulence models.
A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercury
- a) the metal piece will simply float over the mercury
- b) the metal piece will be immersed in mercury by half
- c) whole of the metal piece will be immersed with its top surface just at mercury level
- d) metal piece will sink to the bottom
- e) none of the above
Correct answer is option 'C'. Can you explain this answer?
A piece of metal of specific gravity 13.6 is placed in mercury of specific gravity 13.6, what fraction of it volume is under mercury
a)
the metal piece will simply float over the mercuryb)
the metal piece will be immersed in mercury by halfc)
whole of the metal piece will be immersed with its top surface just at mercury leveld)
metal piece will sink to the bottome)
none of the above|
|
Lekshmi Rane answered |
Ans.
Option (c)
The property of fluid by virtue of which it offers resistance to shear is called- a)surface tension
- b)adhesion
- c)cohesion
- d)viscosity
- e)all of the above
Correct answer is option 'D'. Can you explain this answer?
The property of fluid by virtue of which it offers resistance to shear is called
a)
surface tension
b)
adhesion
c)
cohesion
d)
viscosity
e)
all of the above
|
|
Baishali Bajaj answered |
Viscosity is a property of a fluid by virtue of which it offers resistance to flow. The shear stress at a point in a moving fluid is directly proportional to the rate of shear strain. ... The relationship is known as the Newton's law of viscosity and the fluids which obey this law are known as Newtonian fluids.
Under which of the following conditions will the equation 
constant, will be valid in the whole flow field ?1. Flow is rotational
2. Flow is irrotational
3. Flow is incompressible
4. Flow is steady
5. Flow is laminar- a)1, 3 and 5
- b)2, 4 and 5
- c)1, 3 and 4
- d)2, 3 and 4
- e)Flow is laminar
Correct answer is option 'D'. Can you explain this answer?
Under which of the following conditions will the equation constant, will be valid in the whole flow field ?
constant, will be valid in the whole flow field ?1. Flow is rotational
2. Flow is irrotational
3. Flow is incompressible
4. Flow is steady
5. Flow is laminar
2. Flow is irrotational
3. Flow is incompressible
4. Flow is steady
5. Flow is laminar
a)
1, 3 and 5
b)
2, 4 and 5
c)
1, 3 and 4
d)
2, 3 and 4
e)
Flow is laminar
|
Nitesh Singh answered |
Because of 1-volume doesn't change due to incompressible flow2-Z is constant due to steady flow3-rg doesn't change due to irrotational
For what value of flow behaviour index, does the consistency index has a dimension independent of time?- a)0
- b)1
- c)2
- d)3
Correct answer is option 'C'. Can you explain this answer?
For what value of flow behaviour index, does the consistency index has a dimension independent of time?
a)
0
b)
1
c)
2
d)
3
|
|
Yash Das answered |
Flow Behaviour Index and Consistency Index
Flow behavior index (n) and consistency index (k) are two important parameters used in the power law model which is commonly used to describe non-Newtonian fluids.
The power law model is given by:
τ = k(γ)^n
Where,
τ = shear stress
γ = shear rate
Consistency Index
The consistency index (k) is a measure of the fluid’s resistance to deformation or flow. It is also known as the flow coefficient or the apparent viscosity at low shear rates.
The consistency index has dimensions of Pa s^n or lb s^n/ft^2, where n is the flow behavior index.
Dimension Independence of Time
When the consistency index has dimensions that do not depend on time, it means that the fluid’s behavior is independent of the rate at which the shear stress is applied. This is also referred to as the fluid’s “pseudoplasticity” or “yield stress”. In other words, the fluid’s consistency is not affected by changes in the shear rate or the duration of the shear stress.
Flow Behaviour Index
The flow behavior index (n) is a measure of the degree of non-Newtonian behavior of the fluid. It indicates the extent to which the viscosity of the fluid changes with a change in the shear rate.
A fluid with a flow behavior index of 1 is said to exhibit “Newtonian behavior” because its viscosity remains constant regardless of the shear rate.
Answer
For what value of flow behavior index, does the consistency index have a dimension independent of time?
The correct answer is option 'C' - 2.
When the flow behavior index is equal to 2, the consistency index has dimensions of Pa s^2 or lb s^2/ft^2, which are independent of time.
This means that the fluid’s behavior is not affected by changes in the rate at which the shear stress is applied. Such fluids are referred to as “yield-stress” or “pseudoplastic” fluids.
In summary, when the flow behavior index is equal to 2, the fluid’s consistency is not affected by changes in the shear rate or the duration of the shear stress.
Flow behavior index (n) and consistency index (k) are two important parameters used in the power law model which is commonly used to describe non-Newtonian fluids.
The power law model is given by:
τ = k(γ)^n
Where,
τ = shear stress
γ = shear rate
Consistency Index
The consistency index (k) is a measure of the fluid’s resistance to deformation or flow. It is also known as the flow coefficient or the apparent viscosity at low shear rates.
The consistency index has dimensions of Pa s^n or lb s^n/ft^2, where n is the flow behavior index.
Dimension Independence of Time
When the consistency index has dimensions that do not depend on time, it means that the fluid’s behavior is independent of the rate at which the shear stress is applied. This is also referred to as the fluid’s “pseudoplasticity” or “yield stress”. In other words, the fluid’s consistency is not affected by changes in the shear rate or the duration of the shear stress.
Flow Behaviour Index
The flow behavior index (n) is a measure of the degree of non-Newtonian behavior of the fluid. It indicates the extent to which the viscosity of the fluid changes with a change in the shear rate.
A fluid with a flow behavior index of 1 is said to exhibit “Newtonian behavior” because its viscosity remains constant regardless of the shear rate.
Answer
For what value of flow behavior index, does the consistency index have a dimension independent of time?
The correct answer is option 'C' - 2.
When the flow behavior index is equal to 2, the consistency index has dimensions of Pa s^2 or lb s^2/ft^2, which are independent of time.
This means that the fluid’s behavior is not affected by changes in the rate at which the shear stress is applied. Such fluids are referred to as “yield-stress” or “pseudoplastic” fluids.
In summary, when the flow behavior index is equal to 2, the fluid’s consistency is not affected by changes in the shear rate or the duration of the shear stress.
Newton's law of viscosity is a relationship between- a)shear stress and the rate of angular distortion
- b)shear stress and viscosity
- c)shear stress, velocity and viscosity
- d)pressure, velocity and viscosity
- e)shear stress, pressure and rate of angular distortion.
Correct answer is option 'A'. Can you explain this answer?
Newton's law of viscosity is a relationship between
a)
shear stress and the rate of angular distortion
b)
shear stress and viscosity
c)
shear stress, velocity and viscosity
d)
pressure, velocity and viscosity
e)
shear stress, pressure and rate of angular distortion.
|
|
Rajdeep Gupta answered |
Newton's Law of Viscosity
Newton's Law of Viscosity is a fundamental law that describes the behavior of fluids. It is also known as the law of viscosity or the Newtonian fluid law. This law states that the shear stress of a fluid is directly proportional to the rate of angular distortion of the fluid.
Shear Stress and Rate of Angular Distortion
Shear stress is a measure of the force per unit area that is required to cause a fluid to flow. Rate of angular distortion refers to the rate at which a fluid is deformed when it is subjected to shear stress. The greater the shear stress, the greater the rate of angular distortion.
Relationship between Shear Stress and Rate of Angular Distortion
According to Newton's Law of Viscosity, the relationship between shear stress and the rate of angular distortion is linear. This means that if the shear stress is doubled, the rate of angular distortion will also be doubled. Mathematically, this relationship can be expressed as:
Shear stress = viscosity x rate of angular distortion
where viscosity is a measure of the resistance of a fluid to flow. It is a constant that depends on the nature of the fluid and the temperature at which it is flowing.
Conclusion
In conclusion, Newton's Law of Viscosity is an important law that describes the behavior of fluids. It states that the shear stress of a fluid is directly proportional to the rate of angular distortion of the fluid. This law is applicable to Newtonian fluids, which are fluids that exhibit a linear relationship between shear stress and rate of angular distortion.
Newton's Law of Viscosity is a fundamental law that describes the behavior of fluids. It is also known as the law of viscosity or the Newtonian fluid law. This law states that the shear stress of a fluid is directly proportional to the rate of angular distortion of the fluid.
Shear Stress and Rate of Angular Distortion
Shear stress is a measure of the force per unit area that is required to cause a fluid to flow. Rate of angular distortion refers to the rate at which a fluid is deformed when it is subjected to shear stress. The greater the shear stress, the greater the rate of angular distortion.
Relationship between Shear Stress and Rate of Angular Distortion
According to Newton's Law of Viscosity, the relationship between shear stress and the rate of angular distortion is linear. This means that if the shear stress is doubled, the rate of angular distortion will also be doubled. Mathematically, this relationship can be expressed as:
Shear stress = viscosity x rate of angular distortion
where viscosity is a measure of the resistance of a fluid to flow. It is a constant that depends on the nature of the fluid and the temperature at which it is flowing.
Conclusion
In conclusion, Newton's Law of Viscosity is an important law that describes the behavior of fluids. It states that the shear stress of a fluid is directly proportional to the rate of angular distortion of the fluid. This law is applicable to Newtonian fluids, which are fluids that exhibit a linear relationship between shear stress and rate of angular distortion.
The time oscillation of a floating body with increase in metacentric height will be- a)same
- b)higher
- c)lower
- d)lower/higher depending on weight of body
- e)unpredictable
Correct answer is option 'C'. Can you explain this answer?
The time oscillation of a floating body with increase in metacentric height will be
a)
same
b)
higher
c)
lower
d)
lower/higher depending on weight of body
e)
unpredictable
|
|
Sharmila Chauhan answered |
Explanation:
Metacentric height is a term used in naval architecture and fluid mechanics to describe the distance between the center of gravity of a floating body and its metacentre. The metacentre is the point at which a vertical line through the center of buoyancy of the body intersects with the body in the upright position.
When the metacentric height is increased, it means that the distance between the center of gravity and the metacentre is increased. This has a significant effect on the stability of the floating body and its time oscillation.
Effect of Increase in Metacentric Height:
1. Decreased Stability: When the metacentric height is increased, the stability of the floating body decreases. This is because the higher the metacentric height, the smaller the restoring moment acting on the body. In simple terms, the body becomes less stable and more prone to rolling or tilting.
2. Decreased Time Period of Oscillation: The time period of oscillation refers to the time taken for the floating body to complete one back-and-forth movement. When the metacentric height is increased, the time period of oscillation decreases. This is because the increased distance between the center of gravity and the metacentre reduces the restoring moment acting on the body, causing it to oscillate at a faster rate.
3. Increased Rolling Angle: The rolling angle refers to the angle of tilt or roll experienced by the floating body during oscillation. When the metacentric height is increased, the rolling angle also increases. This is because the increased distance between the center of gravity and the metacentre amplifies the effect of external forces, leading to larger rolling angles.
4. Decreased Damping: Damping refers to the resistance of a system to oscillation or movement. When the metacentric height is increased, the damping of the floating body decreases. This means that the body experiences less resistance to its oscillation, resulting in longer and more pronounced oscillations.
Conclusion:
In conclusion, when the metacentric height of a floating body is increased, its time oscillation is lower. This is due to the decreased stability, reduced time period of oscillation, increased rolling angle, and decreased damping. It is important to consider the metacentric height when designing and analyzing the stability and behavior of floating bodies.
Metacentric height is a term used in naval architecture and fluid mechanics to describe the distance between the center of gravity of a floating body and its metacentre. The metacentre is the point at which a vertical line through the center of buoyancy of the body intersects with the body in the upright position.
When the metacentric height is increased, it means that the distance between the center of gravity and the metacentre is increased. This has a significant effect on the stability of the floating body and its time oscillation.
Effect of Increase in Metacentric Height:
1. Decreased Stability: When the metacentric height is increased, the stability of the floating body decreases. This is because the higher the metacentric height, the smaller the restoring moment acting on the body. In simple terms, the body becomes less stable and more prone to rolling or tilting.
2. Decreased Time Period of Oscillation: The time period of oscillation refers to the time taken for the floating body to complete one back-and-forth movement. When the metacentric height is increased, the time period of oscillation decreases. This is because the increased distance between the center of gravity and the metacentre reduces the restoring moment acting on the body, causing it to oscillate at a faster rate.
3. Increased Rolling Angle: The rolling angle refers to the angle of tilt or roll experienced by the floating body during oscillation. When the metacentric height is increased, the rolling angle also increases. This is because the increased distance between the center of gravity and the metacentre amplifies the effect of external forces, leading to larger rolling angles.
4. Decreased Damping: Damping refers to the resistance of a system to oscillation or movement. When the metacentric height is increased, the damping of the floating body decreases. This means that the body experiences less resistance to its oscillation, resulting in longer and more pronounced oscillations.
Conclusion:
In conclusion, when the metacentric height of a floating body is increased, its time oscillation is lower. This is due to the decreased stability, reduced time period of oscillation, increased rolling angle, and decreased damping. It is important to consider the metacentric height when designing and analyzing the stability and behavior of floating bodies.
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