A piece of pure gold of density $(d_{1} = 9.300 g / {c m}^{3})$ is suspected to be hollow inside. It weighs $38.250 g$ in the air and $33.865 g$ in water. Calculate the volume of the hollow portion in the gold if any.

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Solution

Density of pure gold = $d_{1} = 9.300 g / c m^{2}$
Weight of gold in air = $= 38.250 g$ (in air)
Weight of gold in water = $= 33.865 g$
When it is in air,
mass = $= 38.250 g$
Density $= 9.3 g / c c$
Volume of gold in air $= \frac{m a s s}{d e n s i t y} = \frac{38.250}{9.3} = 4.112 c c$
Loss of mass $= 38.250 - 33.865 = 4.385 g$
According to laws of floatation:-
Mass of water $=$ volume of object
So volume of gold in water = $4.385 c c$
The volume of hollow portion $=$ volume of gold in water $-$ volume of gold in air
$= 4.385 - 4.112 = 0.273 c m^{3}$

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