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For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere as
- a)V1 ∞D1/2 for all diameters
- b)V1 ∞D2 for all diameters
- c)V1 ∞D1/2 for large D and V1 ∞D2 for small D
- d)V1 ∞D2 for large D and V1 ∞D1/2 for small D.
Correct answer is option 'B'. Can you explain this answer?
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For solid spheres falling vertically downwards under gravity in avisco...
Ans. (b) Terminal velocity V1 ∞D2 for all diameters. Stokes’ formula forms the
basis for determination of viscosity of oils which consists of allowing a sphere of known diameter to fail freely in the oil. After initial acceleration. The sphere attains a constant velocity known as Terminal Velocity which is reached when the external drag on the surface and buoyancy, both acting upwards and in opposite to the motions, become equal to the downward force due to gravity.
basis for determination of viscosity of oils which consists of allowing a sphere of known diameter to fail freely in the oil. After initial acceleration. The sphere attains a constant velocity known as Terminal Velocity which is reached when the external drag on the surface and buoyancy, both acting upwards and in opposite to the motions, become equal to the downward force due to gravity.
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For solid spheres falling vertically downwards under gravity in avisco...
Terminal velocity is the maximum velocity attained by an object falling through a fluid under the influence of gravity. It occurs when the drag force acting on the object equals the gravitational force pulling it downwards. The drag force is given by the equation:
Fd = 0.5 * ρ * Cd * A * V^2
Where:
- Fd is the drag force
- ρ is the density of the fluid
- Cd is the drag coefficient
- A is the cross-sectional area of the object
- V is the velocity of the object
The gravitational force is given by:
Fg = m * g
Where:
- Fg is the gravitational force
- m is the mass of the object
- g is the acceleration due to gravity
Since the object is a solid sphere, its mass can be expressed in terms of its density (ρs) and volume (Vs):
m = ρs * V
Where:
- ρs is the density of the sphere
- V is the volume of the sphere
Now, let's consider the variables that affect the terminal velocity:
1. Cross-sectional area (A): The cross-sectional area of a sphere is proportional to the square of its diameter (D^2).
2. Drag coefficient (Cd): The drag coefficient depends on the shape of the object and the Reynolds number, which is a dimensionless parameter related to the fluid flow. For a sphere, the drag coefficient is approximately constant.
3. Density of the fluid (ρ): The density of the fluid remains constant.
4. Density of the sphere (ρs): The density of the sphere remains constant.
From the equations above, we can see that the drag force (Fd) is proportional to the square of the velocity (V^2) and the square of the diameter (D^2). Therefore, we can rewrite the drag force equation as:
Fd = k * D^2 * V^2
Where k is a constant.
Equating the drag force to the gravitational force, we have:
0.5 * ρ * Cd * A * V^2 = ρs * V * g
Substituting the equation for the cross-sectional area (A) and rearranging the terms, we get:
V^2 = (2 * ρs * g) / (ρ * Cd) * (D^2)
Taking the square root of both sides, we get:
V = √[(2 * ρs * g) / (ρ * Cd)] * D
From this equation, we can see that the terminal velocity (V) is directly proportional to the diameter (D). Therefore, the correct answer is option 'B': V1 D2 for all diameters.
Fd = 0.5 * ρ * Cd * A * V^2
Where:
- Fd is the drag force
- ρ is the density of the fluid
- Cd is the drag coefficient
- A is the cross-sectional area of the object
- V is the velocity of the object
The gravitational force is given by:
Fg = m * g
Where:
- Fg is the gravitational force
- m is the mass of the object
- g is the acceleration due to gravity
Since the object is a solid sphere, its mass can be expressed in terms of its density (ρs) and volume (Vs):
m = ρs * V
Where:
- ρs is the density of the sphere
- V is the volume of the sphere
Now, let's consider the variables that affect the terminal velocity:
1. Cross-sectional area (A): The cross-sectional area of a sphere is proportional to the square of its diameter (D^2).
2. Drag coefficient (Cd): The drag coefficient depends on the shape of the object and the Reynolds number, which is a dimensionless parameter related to the fluid flow. For a sphere, the drag coefficient is approximately constant.
3. Density of the fluid (ρ): The density of the fluid remains constant.
4. Density of the sphere (ρs): The density of the sphere remains constant.
From the equations above, we can see that the drag force (Fd) is proportional to the square of the velocity (V^2) and the square of the diameter (D^2). Therefore, we can rewrite the drag force equation as:
Fd = k * D^2 * V^2
Where k is a constant.
Equating the drag force to the gravitational force, we have:
0.5 * ρ * Cd * A * V^2 = ρs * V * g
Substituting the equation for the cross-sectional area (A) and rearranging the terms, we get:
V^2 = (2 * ρs * g) / (ρ * Cd) * (D^2)
Taking the square root of both sides, we get:
V = √[(2 * ρs * g) / (ρ * Cd)] * D
From this equation, we can see that the terminal velocity (V) is directly proportional to the diameter (D). Therefore, the correct answer is option 'B': V1 D2 for all diameters.
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For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer?
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For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer?.
For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For solid spheres falling vertically downwards under gravity in aviscous fluid, the terminal velocity, V1 varies with diameter 'D' of thesphere asa)V1 ∞D1/2for all diametersb)V1 ∞D2for alldiametersc)V1 ∞D1/2forlarge D and V1 ∞D2 for small Dd)V1 ∞D2for large D and V1 ∞D1/2 for small D.Correct answer is option 'B'. Can you explain this answer?.
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