Class 11 Exam > Class 11 Questions > A uniform solid cylinder of density 0.8 g/cm3...
Start Learning for Free
A uniform solid cylinder of density 0.8 g/cm3 floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3 and 1.2 gm/cm3. The height of liquid A is hA = 1.2 cm and the length of the part of cylinder immersed in liquid B is hB = 0.8 cm. Then the length of the cylinder in air is
- a)0.21 m
- b)0.25 cm
- c)0.35 cm
- d)0.4 cm
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in ...
Given:
Density of cylinder (ρc) = 0.8 g/cm³
Density of liquid A (ρA) = 0.7 g/cm³
Density of liquid B (ρB) = 1.2 g/cm³
Height of liquid A (hA) = 1.2 cm
Length of cylinder submerged in liquid B (hB) = 0.8 cm
To find:
Length of the cylinder in air
Analysis:
The cylinder is in equilibrium, which means the density of the cylinder is equal to the average density of the liquid combination. We can calculate the average density using the heights and densities of the liquids.
Let the length of the cylinder in air be hC.
Since the cylinder is in equilibrium, the weight of the cylinder must be equal to the buoyant force acting on it.
Weight of the cylinder = Buoyant force
Weight of the cylinder = Volume of the cylinder × Density of the cylinder × g
Buoyant force = Volume of liquid displaced × Density of the liquid combination × g
The volume of liquid displaced is equal to the sum of the volumes of liquid A and liquid B displaced by the cylinder.
Volume of liquid displaced = Volume of liquid A displaced + Volume of liquid B displaced
Calculations:
Volume of liquid A displaced = Area of the base × height of liquid A
= πr² × hA
Volume of liquid B displaced = Area of the base × length of the cylinder submerged in liquid B
= πr² × hB
Total volume of liquid displaced = πr² × (hA + hB)
Equating the weight of the cylinder to the buoyant force:
πr² × hC × ρc × g = πr² × (hA + hB) × ρavg × g
Canceling out common terms:
hC × ρc = (hA + hB) × ρavg
Substituting the given values:
hC × 0.8 = (1.2 + 0.8) × ρavg
Simplifying:
hC = 2 × ρavg/0.8
The average density (ρavg) is given by:
ρavg = (ρA × hA + ρB × hB)/(hA + hB)
Substituting the given values:
ρavg = (0.7 × 1.2 + 1.2 × 0.8)/(1.2 + 0.8)
Simplifying:
ρavg = (0.84 + 0.96)/2
ρavg = 1.8/2 = 0.9 g/cm³
Substituting the value of ρavg in the equation for hC:
hC = 2 × 0.9/0.8
Simplifying:
hC = 2.25 cm
Therefore, the length of the cylinder in air is 0.25 cm.
Hence, the correct answer is option B.
Density of cylinder (ρc) = 0.8 g/cm³
Density of liquid A (ρA) = 0.7 g/cm³
Density of liquid B (ρB) = 1.2 g/cm³
Height of liquid A (hA) = 1.2 cm
Length of cylinder submerged in liquid B (hB) = 0.8 cm
To find:
Length of the cylinder in air
Analysis:
The cylinder is in equilibrium, which means the density of the cylinder is equal to the average density of the liquid combination. We can calculate the average density using the heights and densities of the liquids.
Let the length of the cylinder in air be hC.
Since the cylinder is in equilibrium, the weight of the cylinder must be equal to the buoyant force acting on it.
Weight of the cylinder = Buoyant force
Weight of the cylinder = Volume of the cylinder × Density of the cylinder × g
Buoyant force = Volume of liquid displaced × Density of the liquid combination × g
The volume of liquid displaced is equal to the sum of the volumes of liquid A and liquid B displaced by the cylinder.
Volume of liquid displaced = Volume of liquid A displaced + Volume of liquid B displaced
Calculations:
Volume of liquid A displaced = Area of the base × height of liquid A
= πr² × hA
Volume of liquid B displaced = Area of the base × length of the cylinder submerged in liquid B
= πr² × hB
Total volume of liquid displaced = πr² × (hA + hB)
Equating the weight of the cylinder to the buoyant force:
πr² × hC × ρc × g = πr² × (hA + hB) × ρavg × g
Canceling out common terms:
hC × ρc = (hA + hB) × ρavg
Substituting the given values:
hC × 0.8 = (1.2 + 0.8) × ρavg
Simplifying:
hC = 2 × ρavg/0.8
The average density (ρavg) is given by:
ρavg = (ρA × hA + ρB × hB)/(hA + hB)
Substituting the given values:
ρavg = (0.7 × 1.2 + 1.2 × 0.8)/(1.2 + 0.8)
Simplifying:
ρavg = (0.84 + 0.96)/2
ρavg = 1.8/2 = 0.9 g/cm³
Substituting the value of ρavg in the equation for hC:
hC = 2 × 0.9/0.8
Simplifying:
hC = 2.25 cm
Therefore, the length of the cylinder in air is 0.25 cm.
Hence, the correct answer is option B.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Top Courses for Class 11View all
A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer?
Question Description
A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect 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 A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect 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 A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer?.
A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect 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 A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect 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 A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A uniform solid cylinder of density 0.8 g/cm3floats in equilibrium in a combination of two non-mixing liquid A and B with its axis vertical. The densities of liquid A and B are 0.7 g/cm3and 1.2 gm/cm3. The height of liquid A is hA= 1.2 cm and the length of the part of cylinder immersed in liquid B is hB= 0.8 cm. Then the length of the cylinder in air isa)0.21 mb)0.25 cmc)0.35 cmd)0.4 cmCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup