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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.

Diagram

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?

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    $\begingroup$ You have to show us what you tried to do, and why it failed. Don't worry about your English! $\endgroup$ – André Chalella Nov 22 '14 at 14:01
  • $\begingroup$ Hint: Archimedean Principle $\endgroup$ – barrycarter Nov 22 '14 at 20:17
  • $\begingroup$ This was a question I was once asked at a job interview $\endgroup$ – user56903 Nov 28 '15 at 16:31
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We know from the fact that the wood block is floating that it is less dense than water. I will assume the coin is more dense than water. I will also assume that we are considering steady state after any waves have died away. I will also assume that the wooden block remains in the same orientation as before. I will also assume the block is cubiod. I will also assume that the density of air is negligable compared to the density of water and so boyancy in air can be ignored.

In the initial state the block and coin are floating together as one "object". So the weight of the water displaced by the block will be equal to the weight of the coin and block together.

After the coin is removed from the block and placed in the water the coin will sink to the bottom. The water displaced by the wooden block will depend on the weight of the wooden block. The water displaced by the coin on the other hand will depend on it's volume (since it has sunk).

The total amount of water displaced will decrease so l will decrease. The wooden block is no longer weighed down by the coin so x will increase.

I'll leave you to try and work out the formula...

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