Question

A piece of gold $(\rho =19.3g/c{m}^3)$ has a cavity in it. It weighs $38.2g$ in air and $36.2g$ in water. The volume of the cavity in gold is

• A. $0.2c{m}^3$
• B. $0.04m^3$
• C. $0.02c{m}^3$
• D. $0.01c{m}^3$

Answer 1

The correct option is C $0.02c{m}^3$

Let the volume of gold piece be $V$ and that of cavity be $v$.

Hence the weight of gold in air=$(V-v)\rho =38.2$

Also weight in water=$(V-v)\rho -V{\rho }_{water}=36.2$

$⟹V=2c{m}^3$

$⟹v=0.02c{m}^3$