
The density of ice is $917{\text{ kg/}}{{\text{m}}^3}$ . What will be the fraction of the volume of a piece of ice above water when it is floating in freshwater?
A. $0.083$
B. $0.042$
C. $0.412$
D. $0.813$
Answer
144.9k+ views
Hint: When a body is partially or completely immersed in a fluid, a force acts on it in the upward direction which is called force of buoyancy or buoyant force. It is given by ${F_B} = \sigma \times V \times g$ where $\sigma $ is the density of the body, $V$ is the volume of the body in consideration and $g$ is the acceleration due to gravity.
Due to this buoyant force, the body loses some of its weight which is equal to the weight of the fluid displaced by the immersed part of the body.
Complete step by step answer
As given in the question, the ice is partially immersed in water.
We know that when a body is partially or completely immersed in a fluid, a force acts on it in the upward direction which is called force of buoyancy or buoyant force. It is given by ${F_B} = \sigma \times V \times g$ where $\sigma $ is the density of the body, $V$ is the volume of the body in consideration and $g$ is the acceleration due to gravity.
And according to the Archimedes’ Principle, due to this buoyant force, the body loses some of its weight which is equal to the weight of the fluid displaced by the immersed part of the body.
Let $f$ be the fraction of volume of the ice that is inside the water. Then $\left( {1 - f} \right)$ will be the fraction of volume above water. We know that the density of water $\rho = 1000{\text{ kg/}}{{\text{m}}^3}$ and the density of ice is given $\sigma = 917{\text{ kg/}}{{\text{m}}^3}$ . Now, we apply Archimedes’ Principle to find the answer i.e.
${\text{Buoyant Force }} = {\text{ Displaced weight of water}}$
$\sigma \times \left( {1 - f} \right) \times V \times g = \rho \times f \times V \times g$
On substituting the value and simplifying we have
\[917 \times \left( {1 - f} \right) = 1000 \times f\]
On further solving we have
$f = 0.917$
Therefore, the fraction of volume of the ice above water is $\left( {1 - f} \right) = 1 - 0.917 = 0.083$
Hence, option A is correct.
Note: The Archimedes’ Principle has numerous important applications in our life. It is used in designing the ships, boats and other water bodies. They are also used in producing Hydrometers which are used to measure density of different liquids.
Due to this buoyant force, the body loses some of its weight which is equal to the weight of the fluid displaced by the immersed part of the body.
Complete step by step answer
As given in the question, the ice is partially immersed in water.
We know that when a body is partially or completely immersed in a fluid, a force acts on it in the upward direction which is called force of buoyancy or buoyant force. It is given by ${F_B} = \sigma \times V \times g$ where $\sigma $ is the density of the body, $V$ is the volume of the body in consideration and $g$ is the acceleration due to gravity.
And according to the Archimedes’ Principle, due to this buoyant force, the body loses some of its weight which is equal to the weight of the fluid displaced by the immersed part of the body.
Let $f$ be the fraction of volume of the ice that is inside the water. Then $\left( {1 - f} \right)$ will be the fraction of volume above water. We know that the density of water $\rho = 1000{\text{ kg/}}{{\text{m}}^3}$ and the density of ice is given $\sigma = 917{\text{ kg/}}{{\text{m}}^3}$ . Now, we apply Archimedes’ Principle to find the answer i.e.
${\text{Buoyant Force }} = {\text{ Displaced weight of water}}$
$\sigma \times \left( {1 - f} \right) \times V \times g = \rho \times f \times V \times g$
On substituting the value and simplifying we have
\[917 \times \left( {1 - f} \right) = 1000 \times f\]
On further solving we have
$f = 0.917$
Therefore, the fraction of volume of the ice above water is $\left( {1 - f} \right) = 1 - 0.917 = 0.083$
Hence, option A is correct.
Note: The Archimedes’ Principle has numerous important applications in our life. It is used in designing the ships, boats and other water bodies. They are also used in producing Hydrometers which are used to measure density of different liquids.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
