Question

A piece of gold with a cavity present in it weighs $10\text{g}$ in air and $9\text{g}$ in water. What is the volume of the cavity?
$[$Density of gold $=19.3{\text{g/cm}}^3]$

• A. $0.182{\text{cm}}^3$
• B. $0.482{\text{cm}}^3$
• C. $1.382{\text{cm}}^3$
• D. $2.482{\text{cm}}^3$

Answer 1

The correct option is B $0.482{\text{cm}}^3$

Loss in weight of the piece of gold is due to the buoyant force acting on it inside water.


$\Rightarrow W_{app}=W_{air}-F_B$
$\Rightarrow W_{app}=W_{air}-\rho g(V_{gold}+V_{cavity})$
$\Rightarrow 9\times {10}^{-3}\times g=10\times {10}^{-3}\times g-{10}^3\times g\times (V_{gold}+V_{cavity})$
$\Rightarrow V_{gold}+V_{cavity}=1\times {10}^{-6}$
$\Rightarrow \frac{m_{g\left(air\right)}}{{\rho }_g}+V_{cavity}=1\times {10}^{-6}$
$\Rightarrow \frac{10\times {10}^{-3}}{19.3\times {10}^3}+V_{cavity}=1\times {10}^{-6}$
$\Rightarrow V_{cavity}=0.482\times {10}^{-6}{\text{m}}^3=0.482{\text{cm}}^3$

$\begin{array}{c}\text{Why this question}?\\ \text{Caution: Please note that while obtaining}{F}_B\\ \text{on gold piece, the total volume of liquid}\\ \text{displaced by it is inclusive of the volume of}\\ \text{cavity.}\end{array}$