Question

You stumble upon a cylindrical block of pure silver with the dimensions radius = 3.45 cm, height = 8.75 cm. If the density of silver is 10.501 g/cm3 and the current value of silver is $16.64/oz, what is the value of the block of silver? (1 lb = 16 oz)

Round your answer to how you would normally round money (to the nearest cent). Do not report your answer in scientific notation.

Answer 1

Given that,
The radius of the cylinder, r = 3.45 cm
The height of the cylinder, h = 8.75 cm

The density of silver,d = 10.501 g/cm3

We know that, the volume of the cylinder, V = πr²h
   = π * (3.45 cm)² * (8.75 cm)
   = 327.021 cm³

By definition,
density = mass/volume.

i.e, d = m/V

10.501 = m/(327.021 cm³)

i.e, mass = 3434.047 g = 7.5707 lb = 121.13 oz

Given that,

The current value of silver is $16.64/oz

Therefore, the value of the block of silver = mass * current value of silver is $16.64/oz

i.e the value of the block of silver = 121.13 * 16.64 = $ 2015.603