A piece of gold `(rho=19.3 g//cm^(3))` has a cavity in it. It weights `38.2g` in air and `36.2g` in water. The volume of the cavity in glold is

Options

A. `0.2cm^(3)`
B. `0.04cm^(3)`
C. `0.02cm^(3)`
D. `0.01cm^(3)`

Answer

C

Explanation

Change in weight=upthurst.
`:. (38.2-36.2)g=(V_(gold)+V_(cavity)) rho_(w)g`
`:. 2=((38.2)/(19.3))+V_(cavity) (as rho_(w)=(1g)/(cm^(3)))`
`:. V_(cavity)=0.02 cm^(3)`.