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The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 is
- a)13.23 cm s-1
- b)17.47 cm s-1
- c)1.47 cm s-1
- d)1.77 cm s-1
Correct answer is option 'B'. Can you explain this answer?
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The terminal velocity of a steel ball of 4 mm in diameter falling thro...
Given:
Diameter of steel ball, d = 4 mm
Relative density of steel, ρs = 8
Relative density of glycerin, ρg = 1.3
Viscosity of glycerin, η = 3.3 poise
To find: Terminal velocity of the steel ball falling through glycerin
Terminal Velocity
Terminal velocity is the maximum velocity achieved by a freely falling object when the resistance of the medium through which it is falling prevents further acceleration.
Stokes' Law
Stokes' law gives the terminal velocity of a small spherical object falling through a viscous fluid. It is given by:
v = (2/9) * [(ρs - ρg) * g * d^2] / η
where,
v = Terminal velocity
ρs = Density of the steel ball
ρg = Density of the glycerin
g = Acceleration due to gravity
d = Diameter of the steel ball
η = Viscosity of the glycerin
Substituting the given values in the above equation, we get:
v = (2/9) * [(8 - 1.3) * 9.8 * (4/1000)^2] / 3.3
v = 0.1747 m/s
v = 17.47 cm/s
Therefore, the terminal velocity of the steel ball falling through glycerin is 17.47 cm/s. Hence, option (b) is the correct answer.
Diameter of steel ball, d = 4 mm
Relative density of steel, ρs = 8
Relative density of glycerin, ρg = 1.3
Viscosity of glycerin, η = 3.3 poise
To find: Terminal velocity of the steel ball falling through glycerin
Terminal Velocity
Terminal velocity is the maximum velocity achieved by a freely falling object when the resistance of the medium through which it is falling prevents further acceleration.
Stokes' Law
Stokes' law gives the terminal velocity of a small spherical object falling through a viscous fluid. It is given by:
v = (2/9) * [(ρs - ρg) * g * d^2] / η
where,
v = Terminal velocity
ρs = Density of the steel ball
ρg = Density of the glycerin
g = Acceleration due to gravity
d = Diameter of the steel ball
η = Viscosity of the glycerin
Substituting the given values in the above equation, we get:
v = (2/9) * [(8 - 1.3) * 9.8 * (4/1000)^2] / 3.3
v = 0.1747 m/s
v = 17.47 cm/s
Therefore, the terminal velocity of the steel ball falling through glycerin is 17.47 cm/s. Hence, option (b) is the correct answer.
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The terminal velocity of a steel ball of 4 mm in diameter falling thro...
= 17.47 cm /s is correct answer
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The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer?
Question Description
The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer?.
The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 isa)13.23 cm s-1b)17.47 cm s-1c)1.47 cm s-1d)1.77 cm s-1Correct answer is option 'B'. Can you explain this answer?.
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