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A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.
- a)0.25
- b)2.5
- c)5
- d)0.5
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on...
The weight of the cubic wooden block is given as 1 kN, which is equivalent to 1000 N (since 1 kN = 1000 N). The inclined plane has an inclination of 30 degrees.
To find the force acting on the block due to gravity, we need to resolve the weight into two components: one parallel to the plane and one perpendicular to the plane.
The weight of the block can be resolved into two components as follows:
Parallel component = weight * sin(angle of inclination)
Perpendicular component = weight * cos(angle of inclination)
Parallel component = 1000 N * sin(30 degrees)
Parallel component = 1000 N * 0.5
Parallel component = 500 N
Perpendicular component = 1000 N * cos(30 degrees)
Perpendicular component = 1000 N * (√3/2)
Perpendicular component = 1000 N * 0.866
Perpendicular component = 866 N
The force acting on the block due to gravity can be calculated by considering the parallel component, which is 500 N.
Therefore, the force acting on the block due to gravity is 500 N.
To find the force acting on the block due to gravity, we need to resolve the weight into two components: one parallel to the plane and one perpendicular to the plane.
The weight of the block can be resolved into two components as follows:
Parallel component = weight * sin(angle of inclination)
Perpendicular component = weight * cos(angle of inclination)
Parallel component = 1000 N * sin(30 degrees)
Parallel component = 1000 N * 0.5
Parallel component = 500 N
Perpendicular component = 1000 N * cos(30 degrees)
Perpendicular component = 1000 N * (√3/2)
Perpendicular component = 1000 N * 0.866
Perpendicular component = 866 N
The force acting on the block due to gravity can be calculated by considering the parallel component, which is 500 N.
Therefore, the force acting on the block due to gravity is 500 N.
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Community Answer
A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on...
Concept:
For a Newtonian fluid,
share stress (τ) = dynamic visocity (μ) × rate of share strain (du/dy)
For the above situation,
where, A = Area in contact of a Newtonian fluid, V = Terminal velocity of the block
Calculation:
Given:
Edge of the block, a = 100 mm = 0.1 m , weight, W = 1 kN, θ = 30°, y = 0.02 mm,
Shear stress acting at the face of the cube in contact with fluid,
V = 5 m/s
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A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. Can you explain this answer?
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A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. 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 cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. 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 cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. Can you explain this answer?.
A cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. 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 cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. 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 cubic wooden block of edge 100 mm and weight 1 kN is sliding down on an inclined plane of inclination 30° with the horizontal. A Newtonian fluid with the viscosity 0.2 Ns/m2 is layered on the inclined plane. If the thickness of the layer is 0.02 mm, then the terminal velocity of the block in m/s.a)0.25b)2.5c)5d)0.5Correct answer is option 'C'. Can you explain this answer?.
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