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

Answer:

0.0195 m

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

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