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 B $0.482 c c$
By using Archimedes principle which states that the buoyant force is equal to the weight of the body.

$m - ρ (V_{g} + V_{c}) = 9 g$

By substituting the value of $V_{g}$ from $m = ρ V_{g}$ in above equation we get,

$m - ρ_{w} \frac{m}{ρ_{g}} - ρ_{w} V_{c} = 9 g$

On solving the above equation we get the value of

$V_{c} = 0.48 c m^{3}$

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