Courses for Kids
Free study material
Offline Centres
Store Icon

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}}$

Last updated date: 14th Jun 2024
Total views: 51.9k
Views today: 0.51k
51.9k+ 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.