A spherical ball has a radius of one inch and a mass of two ounces. Given that the density of water is 0.578 ounces per cubic inch, would the ball float?
In order to answer this question, we firstly need to recall that the density of any object is equal to its mass divided by its volume. We are told that the ball has a mass of two ounces. At present, we don’t know its volume. We do know, however, that the volume of any sphere is equal to four-thirds 𝜋𝑟 cubed. The radius of our ball is equal to one inch. Therefore, 𝑟 is equal to one. The volume of the sphere 𝑉 is therefore equal to four-thirds multiplied by 𝜋 multiplied by one cubed. As one cubed is equal to one, 𝑉 is equal to four-thirds 𝜋. The volume of the spherical ball is therefore four-thirds 𝜋 cubic inches. Volume is always measured in cubic units. And as the radius was in inches, the volume will be in cubic inches.
We can now calculate the density of the ball by dividing the mass by the volume. 𝐷 is equal to two divided by four-thirds 𝜋 or four 𝜋 over three. This simplifies to three over two 𝜋. Typing this into the calculator gives us 0.477464 and so on. If we round this to three decimal places, we get 0.477. The density of the ball is 0.477 ounces per cubic inch.
We can now compare this value to the density of water. This was equal to 0.578 ounces per cubic inch. And 0.477 is less than 0.578. As the density of the ball is less than the density of water, we can say, “Yes, the ball would float.” If, however, our value had been greater than 0.578, the ball would’ve sunk.