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Q.
A piece of gold weighs $10\, g$ in air and $9\, g$ in water. What is the volume of cavity? (Density of gold $=19.3\, g\, cm ^{-3}$)
Mechanical Properties of Fluids
Solution:
$V_{g}, V_{c}$
When dipped in water
$W_{\text {app }}=W_{\text {air }}-F_{B}$
$\Rightarrow 9\, gm \times g=10\, gm \times g-F_{B}$
$\Rightarrow 1 \times g=F_{B}$
Now (total volume displaced) $\times \rho_{w} \times g=1 \times g$
$\left(V_{c}+V_{g}\right) \times 1=1$
$V_{c}=1-\frac{\text { Mass of gold in air }}{\rho_{g}}$
$=1-\frac{10}{19.3}=0.482\, cc$
Where,
$V_{c}=$ volume of cavity
$V_{g}=$ volume of gold
$W_{\text {app }}=9\, gm$
$W_{\text {air}}=10\, gm$
$F_{B}=$ force of buoyancy
$\rho_{w}=$ density of water $=1$
$\rho_{g}=$ density of gold $=19.3$