Question

A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a height of 2.8 cm. If this puck is placed into a bowl of mercury (density 13.6 grams per cubic centimeter), it floats. How deep below the surface of the mercury is the bottom of the metal puck?

Answer 1

0.0195 m

Step-by-step explanation:

`\rho _(p)` = density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³

`d_(p)` = diameter of hockey puck = 13 cm = 0.13 m

`h_(p)` = height of hockey puck = 2.8 cm = 0.028 m

`\rho _(m)` = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

`d` = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

`\rho _(m) (0.25)(\pi d_(p)^(2) d ) g = \rho _(p) (0.25)(\pi d_(p)^(2) h_(p) ) g\\\rho _(m) ( d ) = \rho _(p) ( h_(p) )\\(13600) d = (9450) (0.028)\\d = 0.0195 m`