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A piece of gold weighs $10 \,g$ in air and $9 \,g$ in water. What is the volume of cavity is ...... $cc$ (Density of gold $=19.3 \,g cm ^{-3}$ )
$0.182$
$0.282$
$0.382$
$0.482$
Solution
(d)
When dipped in water
$W_{\text {app }}=W_{\text {air }}-F_B$
$\Rightarrow 9 \,gm \times g=10 \,gm \times g-F_B$
Where,
$\left\{\begin{array}{l}V_c=\text { volume of cavity } \\ V_g=\text { volume of gold } \\ W_{\text {app }}=9 gm \\ W_{\text {air }}=10 gm \\ F_B=\text { force of buoyancy } \\ \rho_w=\text { density of water }=1 \\ \rho_g=\text { density of gold }=19.3\end{array}\right.$
$\Rightarrow 1 \times g=F_B$
Now (total volume displaced) $\times \rho_w \times g=1 \times g$
$\left(V_c+V_g\right) \times 1=1$
$V_c=1-\frac{\text { Mass of gold in air }}{\rho_g}=1-\frac{10}{19.3}=0.482 \,cc$
