Question

Density of ice is $900kg/{m^3}$. A piece of ice is floating in water (density = $1000kg/{m^3}$). The fraction of volume of the piece of ice outside the water is:
A. $0.9$
B. $0.1$
C. $0$
D. $0.7$

Answer 1

We know that buoyancy force is an upward force which is exerted by any fluid upon a body placed on it. Because of this force objects float in the fluids or we can say that when any body is submerged in a fluid or liquid then it becomes lighter because of some downward force of the weight of the body is balanced by the upthrust of the liquid which is known as upward buoyancy force of the liquid.

Complete step-by-step answer:
If the total volume of the ice is $V$ and ${V_1}$ is the volume of ice piece immersed in water, Thus at the equilibrium the weight of the ice and upthrust force or buoyant force will be equal.
Now weight of ice = buoyant force
$\Rightarrow V{\rho _i}g = {V_1}{\rho _w}g$ ; Here ${\rho _i}$ is density of ice and ${\rho _w}$ is density of water.
Now we will put the values of ${\rho _i}$ and ${\rho _w}$ which is given in the problem statement.
$\Rightarrow V{\rho _i} = {V_1}{\rho _w} \\\ \Rightarrow V \times 900 = {V_1} \times 1000 \\\ \Rightarrow \dfrac{{{V_1}}}{V} = \dfrac{{900}}{{1000}} = 0.9 \\\$
Now we know that in the problem it is asked the value of the fraction of volume of the piece of ice outside the water so $fraction = 1 - 0.9 = 0.1$
Therefore option B is the correct answer to this problem which is the fraction of volume of the piece of ice outside the water is $0.1$.

So, the correct answer is “Option B”.

Note: We have solved this problem by equating the weight of ice to the upthrust of liquid because when both will be equal at equilibrium only then ice would float in water. At last we have calculated the fraction of volume of the piece of ice which is outside the water.