# Help Zone

### Student Question

Secondary III • 1yr.

Someone throws a huge 300 kg object into a 40,000 L swimming pool without any splashes being produced. The pool’s volume is measured afterwards, thus showing that it reaches 45,000 L. What is the density of the object? Help !?!

Science

## Explanations (1)

• Explanation from Alloprof

Explanation from Alloprof

This Explanation was submitted by a member of the Alloprof team.

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Team Alloprof • 1yr.

Let's first start by compiling what we know and what we can deduce from the question:

• The pool’s initial volume is 40,000 L

• The pool’s final volume is 45,000 L

• The object’s has a mass of 300 kg

Next, we can define what we are looking for:

• The density of the object

To begin with, we must find the variation in volume in the pool. Indeed, a volume of water was already present in the pool and it shouldn’t be considered as being part of the volume of the object. Thus, it is necessary to carry out the following calculation:

Volume objet = volume final-volume initial

$$volume\:objet=45000\:L-40000\:L=5000\:L$$

Then, it is necessary to calculate the density of the object with the formula for density:

$$\rho=\frac{m}{V}$$

However, the units are not appropriate for the situation. In general, density is expressed in g / mL or g / cm ^ 3. Thus, the object’s volume must be converted into mL:

$$5000\:L = 5000\:L • \frac{1000\:mL}{L}=5\:000\:000\:mL$$

The object’s mass must then be converted into g:

$$300\:kg=300\:kg•\frac{1000\:g}{kg}=300\:000\:g$$

The final calculation is as follows:

$$\rho=\frac{300\:000\:g}{5\:000\:000\:mL}=0,06\:\frac{g}{mL}$$

There you go!

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