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The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.
(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).
- a)0.116
- b)0.104
- c)0.119
- d)0.107
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The maximum size of sand particles (with relative density of 2.65) wh...
Stoke’s law is valid up to
Re = 1.0
For maximum size particle that obey stoke’s law,
Also the setting velocity,
Substituting for Vt
Most Upvoted Answer
The maximum size of sand particles (with relative density of 2.65) wh...
Stoke's Law:
Stoke's law is used to calculate the settling velocity of small spherical particles in a viscous fluid. It states that the settling velocity of a particle is directly proportional to the square of its radius and the difference in density between the particle and the fluid, and inversely proportional to the dynamic viscosity of the fluid.
Formula:
The formula for the settling velocity of a particle according to Stoke's law is:
V = (2/9) * (g * (d^2) * (ρp - ρf)) / η
where,
V is the settling velocity of the particle,
g is the acceleration due to gravity (9.81 m/s^2),
d is the diameter of the particle,
ρp is the density of the particle,
ρf is the density of the fluid (water),
η is the dynamic viscosity of the fluid.
Calculating the maximum size of sand particles:
In this question, we are given the relative density of the sand particles (ρp/ρf = 2.65), the density of water (ρf = 998 kg/m^3), and the dynamic viscosity of water (η = 1 × 10^-3 Pa.s).
We need to find the maximum size of sand particles, which means we need to find the maximum diameter (d) that will allow the particles to settle in water.
To find the maximum diameter, we can rearrange the Stoke's law formula as follows:
d = √((9 * V * η) / (2 * g * (ρp - ρf)))
Substituting the given values into the formula:
d = √((9 * 0.02 * 1 × 10^-3) / (2 * 9.81 * (2.65 - 998)))
d ≈ 0.104 mm
Therefore, the maximum size of sand particles that will settle in water, according to Stoke's law, is approximately 0.104 mm.
Answer:
Option 'B' (0.104 mm) is the correct answer.
Stoke's law is used to calculate the settling velocity of small spherical particles in a viscous fluid. It states that the settling velocity of a particle is directly proportional to the square of its radius and the difference in density between the particle and the fluid, and inversely proportional to the dynamic viscosity of the fluid.
Formula:
The formula for the settling velocity of a particle according to Stoke's law is:
V = (2/9) * (g * (d^2) * (ρp - ρf)) / η
where,
V is the settling velocity of the particle,
g is the acceleration due to gravity (9.81 m/s^2),
d is the diameter of the particle,
ρp is the density of the particle,
ρf is the density of the fluid (water),
η is the dynamic viscosity of the fluid.
Calculating the maximum size of sand particles:
In this question, we are given the relative density of the sand particles (ρp/ρf = 2.65), the density of water (ρf = 998 kg/m^3), and the dynamic viscosity of water (η = 1 × 10^-3 Pa.s).
We need to find the maximum size of sand particles, which means we need to find the maximum diameter (d) that will allow the particles to settle in water.
To find the maximum diameter, we can rearrange the Stoke's law formula as follows:
d = √((9 * V * η) / (2 * g * (ρp - ρf)))
Substituting the given values into the formula:
d = √((9 * 0.02 * 1 × 10^-3) / (2 * 9.81 * (2.65 - 998)))
d ≈ 0.104 mm
Therefore, the maximum size of sand particles that will settle in water, according to Stoke's law, is approximately 0.104 mm.
Answer:
Option 'B' (0.104 mm) is the correct answer.
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The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer?
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The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer?.
The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The maximum size of sand particles (with relative density of 2.65) which will settle in water, according to Stoke’s law, is ___________mm.(Density of water = 998kg/m3 , Dynamic viscosity = 1×10-3 Pa.s).a)0.116b)0.104c)0.119d)0.107Correct answer is option 'B'. Can you explain this answer?.
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