A piece of gold weighs $10 g$ in air and $9 g$ in water. What is the volume of cavity?
(density of gold $= 19.3 g {c m}^{- 3}$ )

0.182 c c

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0.282 c c

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0.382 c c

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0.482 c c

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Solution

The correct option is D $0.482 c c$
For piece of gold in air
$m = v_{g} ρ_{g} = 10 g (1)$
According to Archemedes Principal
$m - ρ_{w} (v_{g} - v_{e}) = 9 g (2)$
Where
$v_{e} =$ volume of cavity
On solving these two equations, we get,
$m - ρ_{w} (\frac{m}{ρ_{g}}) - ρ_{w} v_{e} = 9 v_{e} = \frac{10 (1 - \frac{ρ_{w}}{ρ_{g}}) - 9}{ρ_{w}} = 0.48 c m^{3}$

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