
An Ice cube of size $a = 10$cm is floating in a tank (base area A $ = 50 \times 50$ $c{m^2}$) partially filled with water. The change in gravitational potential energy when ice melts completely is (density of ice is $900$ kg/ ${m^3}$ )
A) $- 0.072{\text{J}}$
B) $ - {\text{0}}{\text{.24J}}$
C) $ - 0.016{\text{J}}$
D) $ - 0.0045{\text{J}}$
Answer
133.8k+ views
Hint: The basic approach to solve this question is by the use of mass density formulas, potential energy in terms of density and the concept used is mass conservation as the ice melts its mass remains the same which got added to the initial mass of water, the only change occur is in the total energy in single state that means Total energy before and the process remains same.
Complete step by step answer:
First let us study the concept we will going to use in the solution, Here we will use the concept of potential energy but in moulded form that means in the formula of Potential Energy we will convert the mass in form of density, length and area and Secondly will use the fraction parts to deal with ice-water and water mixture, So let us start with fraction parts,
According to the question given and on the basis of the given values we have some calculations regarding change in densities of ice when it is partially filled in water.
volume of the ice cube $ = $ 0.001 ${m^3}$ so the mass of the ice is $0.001 \times 900 = 0.9$ kg
after melting, the Ice become water then the volume will be $\dfrac{{0.9}}{{1000}} = 0.0009$ kg
So, the extra volume of the ice cube that was above the water in the container is $0.001 - 0.0009 = 0.0001\,\,{m^3}$
after the ice becomes water it will fit inside the water volume hence the change in the potential energy for the portion that was outside from the water. The height h of the portion is $\dfrac{{0.0001}}{{0.001}} = 0.01$ m
Now the potential energy for a slab is $P = A\rho g\dfrac{{{h^2}}}{2}$
(where $A$ is the area $h$ is the height and $\rho $ is the density)
Putting the values, we get
$\rho = 0.001 \times 900 \times 9.8 \times \dfrac{{{{0.01}^2}}}{2} = 0.0045$
Hence potential energy will be decreased by 0.0045 J
So correct option is (D)
Note: If the ice is fully merged then we will simplify the total potential energy before and after because in this case total or full mass is inside the water and followed by this can simply calculate the increase in height.
Complete step by step answer:
First let us study the concept we will going to use in the solution, Here we will use the concept of potential energy but in moulded form that means in the formula of Potential Energy we will convert the mass in form of density, length and area and Secondly will use the fraction parts to deal with ice-water and water mixture, So let us start with fraction parts,
According to the question given and on the basis of the given values we have some calculations regarding change in densities of ice when it is partially filled in water.
volume of the ice cube $ = $ 0.001 ${m^3}$ so the mass of the ice is $0.001 \times 900 = 0.9$ kg
after melting, the Ice become water then the volume will be $\dfrac{{0.9}}{{1000}} = 0.0009$ kg
So, the extra volume of the ice cube that was above the water in the container is $0.001 - 0.0009 = 0.0001\,\,{m^3}$
after the ice becomes water it will fit inside the water volume hence the change in the potential energy for the portion that was outside from the water. The height h of the portion is $\dfrac{{0.0001}}{{0.001}} = 0.01$ m
Now the potential energy for a slab is $P = A\rho g\dfrac{{{h^2}}}{2}$
(where $A$ is the area $h$ is the height and $\rho $ is the density)
Putting the values, we get
$\rho = 0.001 \times 900 \times 9.8 \times \dfrac{{{{0.01}^2}}}{2} = 0.0045$
Hence potential energy will be decreased by 0.0045 J
So correct option is (D)
Note: If the ice is fully merged then we will simplify the total potential energy before and after because in this case total or full mass is inside the water and followed by this can simply calculate the increase in height.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Class 11 JEE Main Physics Mock Test 2025

Current Loop as Magnetic Dipole and Its Derivation 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

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
