Question

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

• A. 0.182 cc
• B. 0.282 cc
• C. 0.382 cc
• D. 0.482 cc

Answer 1

Answer: (D)

$V_g,V_c$
When dipped in water
$W_{\text{app }}=W_{\text{air }}-F_B$
$\Rightarrow 9gm\times g=10gm\times g-F_B$
$\Rightarrow 1\times g=F_B$
Now (total volume displaced) $\times {\rho }_w\times g=1\times g$
$\left(V_c+V_g\right)\times 1=1$
$V_c=1-\frac{\text{ Mass of gold in air }}{{\rho }_g}$
$=1-\frac{10}{19.3}=0.482cc$
Where,
$V_c=$ volume of cavity
$V_g=$ volume of gold
$W_{\text{app }}=9gm$
$W_{\text{air}}=10gm$
$F_B=$ force of buoyancy
${\rho }_w=$ density of water $=1$
${\rho }_g=$ density of gold $=19.3$