Question

A piece of gold weights $10g$ in air and $9g$ in water. What is the volume of cavity? Density of gold $=19.3gm/c{m}^3$.

• A. None
• B. $0.482cc$
• C. $0.282cc$
• D. $0.682cc$

Answer 1

The correct option is B $0.482cc$

Given, weight of gold piece in air,

$W_0=10g$

Weight of gold piece in water,

$W_a=9g$

Density of the gold, $\sigma =19.3gm/c{m}^3$

Actual volume of gold $=\frac{10}{19.3}$

$=0.518c{m}^3$

Now, if $V$ be the apparent volume of the piece of gold,

Then, volume of the displaced liquid,

$V=\frac{W_0-W_a}{\rho g}$

$\Rightarrow V=\frac{(10-9)g}{1g}=1c{m}^3$

So, volume of cavity $=1-0.518$

$=0.482cc$

Final answer : (b)