# Metallic coin over wooden block in glass of water

So there is some water inside a container; the height of water inside the container is $l$. I placed a wooden block on the water and it's floating to some height $x$, on top of the block is a metallic coin (see the diagram below).

What will happen if I throw that coin inside the water? Will the Height $x$ and $l$ change? Can we derive an expression for this change? I tried doing it, but failed.

Edited : Here is what I've tried Let's assume that our wooden block has mass $m$1 Volume $v$1height $h$1, height underneath the water as $y$ and let's assume mass of Metallic coin as $m$2 Now total mass on block will be equal to (in which buoyant force is actin upon) : $m$ $=$ $m$2 + $m$1 and let's also say that total height $h$1 = $x$ + $y$

Principle of buoyancy of a liquid is Give as : $F$b$= gρhA$
Simplifying :
$F$b$= g(m/v1)(x+y)A$ ... {1}

Now, archimedes principle is given as $F = gρhA$ $F = g(ρ$f - $ρ$b$)hA$ // $ρ$b is Density of the Body

And then I am not sure what to do next .. Can anyone help me out?

• You have to show us what you tried to do, and why it failed. Don't worry about your English! Commented Nov 22, 2014 at 14:01
• Hint: Archimedean Principle
– user854
Commented Nov 22, 2014 at 20:17
• This was a question I was once asked at a job interview
– user56903
Commented Nov 28, 2015 at 16:31