# A spherical vessel (diameter #"= 2.00 cm"#) when empty has a mass of #"2.00 g"#. What is the greatest volume of water that can be placed in the vessel and still have the vessel float at the surface of the water?

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Given: density of water #= "1.00 g/cm"^3#

Given: density of water

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

For starters, you know that in order for an object to **float** on water, i.e. at the surface of the water, its **density** must be **lower** than the density of water.

In other words, the ratio that exists between the **mass** of the object and the **volume** it occupies must be **lower** than

So, start by calculating the total volume of the sphere. You know

#V_"sphere" = 4/3 * pi * r^3#

Here

#r# is theradiusof the sphere, equal tohalfitsdiameter,#d#

In your case, you have

#d = "2.00 cm"#

which means that

#r = "2.00 cm"/2 = "1.00 cm"#

The volume of the sphere will thus be equal to

#V_"sphere" = 4/3 * pi * "1.00 cm"^3#

#V_"sphere" = "4.189 cm"^3#

Now, you know that the **empty sphere** has a mass of **density** when **empty** is equal to

#rho_"empty" = "2.00 g"/"4.189 cm"^3 = "0.477 g cm"^(-3)#

Since the density of the empty sphere is **lower** than **float**.

At this point, your goal is to figure out how much water will **increase** the **total mass** of the sphere to the point

#"mass sphere"/"4.189 cm"^3 <= "1.00 g cm"^(-3)#

Since you want the sphere to float *at the surface* of the water, you can take the **maximum value** of that expression and say that you need

#"mass sphere"/"4.189 cm"^3 = "1.00 g cm"^(-3)# Keep in mind that technically, this is not the maximum value that the density of the sphere can take in order for it to float

on the surfaceof the water because when the two densities areequal, the sphere will besuspendedin the water.

This is equivalent to saying that the mass of the sphere must be equal to

#"mass sphere" = 4.189 color(red)(cancel(color(black)("cm"^3))) * "1.00 g" color(red)(cancel(color(black)("cm"^(-3))))#

#"mass sphere = 4.189 g"#

But remember, this includes the mass of the added water **and** the mass of the sphere itself, so you will have

#"mass added water" = "4.189 g " - " 2.00 g"#

#"mass added water = 2.189 g"#

Finally, to convert the mass of water to *cubic centimeters*, use the given density of water.

#2.189 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(1.00color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("2.19 cm"^3)))#

The answer is rounded to three **sig figs**.

So one could argue that something like **suspended** in the water, i.e. it will not float *on the surface*.

For example, in a multiple-choice problem that has