A piece of steel has a weight `w` in air, `w_(1)` when completely immersed in water and `w_(2)` when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid is

Options

A. `(w-w_(1))/(w-w_(2))`
B. `(w-w_(2))/(w-w_(1))`
C. `(w_(1)-w_(2))/(w-w_(1))`
D. `(w_(1)-w_(2))/(w-w_(2))`

Answer

B

Explanation

Relative density of steel=`("weight in air")/("change in weight in water")`
`=(w)/(w-w_(1))`
Now change in weight in the given liquid = upthrust in this liquid
or `w-w_(2)={(w//g)/((w//w-w_(1))rho_(w))}d_(2)xxrho_(w)xxg`
`therefore d_(2)=(w-w_(2))/(w-w_(1))`= relative density of given liquid.