Question

Someone tries to sell you a solid but irregularly shaped lump of gold, but you are suspicious because you think that it feels too light. Because it is irregularly shaped, you cannot easily calculate its volume from measurable linear dimensions. So this is what you do. You weigh the lump of gold on a pan balance (weight scale) that reads in kg up to a maximum of 80 kg. The scale reading is 25.95 kg with the gold lump resting on the pan of the scale. You remove the gold lump from the scale and is replaced it with a beaker of mercury (a liquid metal) which has a density of 1.36 Times 10^4 kg/m^3; the scale now reads 50.00 kg. Ignoring the poisonous vapor rising from the mercury, you suspend lump of gold from a thread as shown in the diagram, and lowered it into the mercury until it is completely immersed but does not touch the bottom of the beaker. The reading of the balance increases from 50.00 kg to 71.36 kg when the lump of gold is immersed in the liquid. The density of gold is l.93 Times 104 kg/m3. A08. What is the upthrust (buoyancy force) exerted on the hollow lump of gold when it is totally immersed in the liquid? What is the average density of the supposedly solid lump of gold? What is the volume of the hollow space inside the supposedly solid lump of gold?

Answer 1

8) Upthrust = Reduction is weight x g = (25.95 - (71.36-50))g=4.59g=44.982 N

9) Volume of solid = Upthrust/ Density of Water = 44.982/1000g=0.00459 m^3

   density = m/V=25.95/0.00459=5653.59 kg/m^3

10) .Avg Density = Density of Gold x (Total Volume - Hollow Volume)/Total Volume (Neglecting the density of hollow space)

Hollow Volume= (1-(Avg Density/Density of Gold))x Total VOlume= 0.00324 m^3