# A pure water ice cube with sides 2cm long floats in a liquid in a glass beaker [duplicate]

A pure water ice cube with sides $2\,\mathrm{cm}$ long floats in a liquid in a glass beaker.

What volume will be below the liquid surface if the liquid is

a) pure water and

b) brine?

What happens to the level of the water in the beaker if the ice cube melts and why?

The density of pure water is $1000\,\mathrm{kgm^{-3}}$

The density of brine is $1150\,\mathrm{kgm^{-3}}$

(Both given to me from my question sheet).

I'm doing AS physics and I don't know even how to answer this question, all help will be greatly appreciated. (I would also be very grateful if you try to make it as uncomplicated as possible).

My attempt to answer the first question was by finding the volume of the ice cube which is $8 \times 10^{-6}$ but I wasn't sure what to do with that. Or if I even needed that!

For the second question, I believe the answer goes along the lines of "there is no increase to the level of water in the beaker as the ice cube already displaced an amount of fluid once it was placed in it." Is there anything I should add to that?

Archimedes' Principle states that an object will displace a volume of fluid equal to its weight. You need to know the density of ice to answer this question (if it was the same as the density of the water the ice would not float) which is approximately $\rho_{\textrm{ice}} = 920 \textrm{ kgm}^{-3}$.
The mass of the ice cube is $\rho V$ which is $920 \times 8 \times 10^{-6} = 7.36 \times 10^{-3} \textrm{ kg}$ or 7.36 grams. The weight of 7.36 grams is 72.2 mN, which means that your ice cube will displace 72.2 mN of water.
Given the density of water as $\rho_{\textrm{water}} = 1000 \textrm{ kgm}^{-3}$ this means that the volume of water displaced (and therefore the volume of the ice submerged) is $\frac{72.2 \times 10^{-3}}{1000 \times 9.81} = 7.36 \times 10^{-6} \textrm{ m}^3$. You can repeat this for brine using the density you provided.