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A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed of
- a)0.5 cm/s
- b)13.5 cm/s
- c)1.5 cm/s
- d)4.5 cm/s
Correct answer is option 'D'. Can you explain this answer?
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A Newtorian fluid fills the clearance between a shaft and a sleeve, wh...
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A Newtorian fluid fills the clearance between a shaft and a sleeve, wh...
To solve this problem, we need to apply the concept of Newtonian fluids and the relationship between force and velocity.
Newtonian Fluids:
A Newtonian fluid is a type of fluid in which the shear stress is directly proportional to the rate of shear strain. In simpler terms, the force required to move a Newtonian fluid at a certain velocity is directly proportional to that velocity.
Relationship between Force and Velocity:
According to Newton's second law of motion, the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force is parallel to the sleeve and is causing the shaft to move.
Given Information:
Force applied to the shaft (F1) = 0.8 kN
Velocity of the shaft (V1) = 1.5 cm/s
Using the relationship between force and velocity, we can set up a proportion to find the force required to achieve a different velocity.
F1/V1 = F2/V2
Where F2 is the force required to achieve a different velocity (V2).
Now, we can substitute the given values into the equation:
0.8 kN / 1.5 cm/s = F2 / V2
To find the force required for a different velocity, we need to solve for F2.
F2 = (0.8 kN / 1.5 cm/s) * V2
Given that the force is now 2.4 kN, we can substitute this value into the equation and solve for V2.
2.4 kN = (0.8 kN / 1.5 cm/s) * V2
Dividing both sides by 0.8 kN / 1.5 cm/s, we get:
V2 = (2.4 kN) / (0.8 kN / 1.5 cm/s)
Simplifying further:
V2 = 3 cm/s * 1.5 cm/s
V2 = 4.5 cm/s
Therefore, when a force of 2.4 kN is applied to the shaft, parallel to the sleeve, the shaft would move with a speed of 4.5 cm/s, which corresponds to option D.
Newtonian Fluids:
A Newtonian fluid is a type of fluid in which the shear stress is directly proportional to the rate of shear strain. In simpler terms, the force required to move a Newtonian fluid at a certain velocity is directly proportional to that velocity.
Relationship between Force and Velocity:
According to Newton's second law of motion, the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force is parallel to the sleeve and is causing the shaft to move.
Given Information:
Force applied to the shaft (F1) = 0.8 kN
Velocity of the shaft (V1) = 1.5 cm/s
Using the relationship between force and velocity, we can set up a proportion to find the force required to achieve a different velocity.
F1/V1 = F2/V2
Where F2 is the force required to achieve a different velocity (V2).
Now, we can substitute the given values into the equation:
0.8 kN / 1.5 cm/s = F2 / V2
To find the force required for a different velocity, we need to solve for F2.
F2 = (0.8 kN / 1.5 cm/s) * V2
Given that the force is now 2.4 kN, we can substitute this value into the equation and solve for V2.
2.4 kN = (0.8 kN / 1.5 cm/s) * V2
Dividing both sides by 0.8 kN / 1.5 cm/s, we get:
V2 = (2.4 kN) / (0.8 kN / 1.5 cm/s)
Simplifying further:
V2 = 3 cm/s * 1.5 cm/s
V2 = 4.5 cm/s
Therefore, when a force of 2.4 kN is applied to the shaft, parallel to the sleeve, the shaft would move with a speed of 4.5 cm/s, which corresponds to option D.
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A Newtorian fluid fills the clearance between a shaft and a sleeve, wh...
Given data:
Force applied to the shaft, F1 = 0.8 kN
Speed of the shaft, V1 = 1.5 cm/s
To find:
The speed of the shaft when a force of 2.4 kN is applied, V2.
Assumption:
The fluid between the shaft and sleeve is a Newtonian fluid.
Explanation:
Newton's law of viscosity states that the shear stress (τ) between two layers of a fluid is directly proportional to the velocity gradient (dv/dy) and the viscosity (µ) of the fluid. Mathematically, it can be represented as:
τ = µ(dv/dy)
For a Newtonian fluid, the shear stress is linearly proportional to the velocity gradient.
Using the given data:
Force F1 = 0.8 kN = 800 N
Speed V1 = 1.5 cm/s
Calculating shear stress and velocity gradient:
Shear stress (τ) = Force (F1) / Area (A)
The area (A) can be calculated using the clearance between the shaft and sleeve.
Assuming the clearance between the shaft and sleeve is 'd',
Area (A) = πd^2 / 4
Shear stress (τ) = F1 / (πd^2 / 4)
Velocity gradient (dv/dy) = V1 / d
Using Newton's law of viscosity:
τ = µ(dv/dy)
µ(dv/dy) = F1 / (πd^2 / 4)
µ = (4F1) / (πd^2V1)
Now, let's calculate the viscosity (µ) using the given data.
Calculating viscosity (µ):
µ = (4 * 800) / (π * d^2 * 1.5)
Now, let's calculate the speed of the shaft when a force of 2.4 kN is applied.
Using the calculated viscosity (µ):
Force F2 = 2.4 kN = 2400 N
Shear stress (τ) = F2 / (πd^2 / 4)
Velocity gradient (dv/dy) = V2 / d (to be calculated)
τ = µ(dv/dy)
µ(dv/dy) = F2 / (πd^2 / 4)
Now, substituting the values and solving for V2:
µ * (V2 / d) = F2 / (πd^2 / 4)
V2 = (F2 * 4 * d) / (πd^2 * µ)
Substituting the values of F2, d, and µ:
V2 = (2400 * 4 * d) / (πd^2 * [(4 * 800) / (π * d^2 * 1.5)])
V2 = (2400 * 4 * d * (π * d^2 * 1.5)) / (πd^2 * 4 * 800)
V2 = (2400 * 6 * d^3) / (4 * 800)
V2 = (2400 * 6
Force applied to the shaft, F1 = 0.8 kN
Speed of the shaft, V1 = 1.5 cm/s
To find:
The speed of the shaft when a force of 2.4 kN is applied, V2.
Assumption:
The fluid between the shaft and sleeve is a Newtonian fluid.
Explanation:
Newton's law of viscosity states that the shear stress (τ) between two layers of a fluid is directly proportional to the velocity gradient (dv/dy) and the viscosity (µ) of the fluid. Mathematically, it can be represented as:
τ = µ(dv/dy)
For a Newtonian fluid, the shear stress is linearly proportional to the velocity gradient.
Using the given data:
Force F1 = 0.8 kN = 800 N
Speed V1 = 1.5 cm/s
Calculating shear stress and velocity gradient:
Shear stress (τ) = Force (F1) / Area (A)
The area (A) can be calculated using the clearance between the shaft and sleeve.
Assuming the clearance between the shaft and sleeve is 'd',
Area (A) = πd^2 / 4
Shear stress (τ) = F1 / (πd^2 / 4)
Velocity gradient (dv/dy) = V1 / d
Using Newton's law of viscosity:
τ = µ(dv/dy)
µ(dv/dy) = F1 / (πd^2 / 4)
µ = (4F1) / (πd^2V1)
Now, let's calculate the viscosity (µ) using the given data.
Calculating viscosity (µ):
µ = (4 * 800) / (π * d^2 * 1.5)
Now, let's calculate the speed of the shaft when a force of 2.4 kN is applied.
Using the calculated viscosity (µ):
Force F2 = 2.4 kN = 2400 N
Shear stress (τ) = F2 / (πd^2 / 4)
Velocity gradient (dv/dy) = V2 / d (to be calculated)
τ = µ(dv/dy)
µ(dv/dy) = F2 / (πd^2 / 4)
Now, substituting the values and solving for V2:
µ * (V2 / d) = F2 / (πd^2 / 4)
V2 = (F2 * 4 * d) / (πd^2 * µ)
Substituting the values of F2, d, and µ:
V2 = (2400 * 4 * d) / (πd^2 * [(4 * 800) / (π * d^2 * 1.5)])
V2 = (2400 * 4 * d * (π * d^2 * 1.5)) / (πd^2 * 4 * 800)
V2 = (2400 * 6 * d^3) / (4 * 800)
V2 = (2400 * 6
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A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer?
Question Description
A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer?.
A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Newtorian fluid fills the clearance between a shaft and a sleeve, when a force of 0.8 kN is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 1.5 cm/s. If a force of 2.4 kN is applied instead the shaft would move with a speed ofa)0.5 cm/sb)13.5 cm/sc)1.5 cm/sd)4.5 cm/sCorrect answer is option 'D'. Can you explain this answer?.
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