Question

A piece of metal weighs $50.0N$ in air, $36.0N$ in water, and $41.0N$ in an unknown liquid. Find the densities the metal.

• A. $3.57\times {10}^{3}kg/m^3$
• B. $3.57kg/m^3$
• C. $1.389\times {10}^{3}kg/m^3$
• D. $0.72kg/m^3$

Answer 1

The correct option is A $3.57\times {10}^{3}kg/m^3$

In this case , firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = $50N$ = $mg$ implies mass of metal is $5kg$

Now difference of weight of metal in air and water = upthrust acting on it = $volum{e}_{\left(metal\right)}{\rho }_{(}liquid)g=V(1000\left)\right(10)=14N$

So volume of metal piece = $14\times {10}^{-4}kg/m^3$

So density of metal = mass of metal / volume of metal = $5/14\times {10}^{-4}$ = $3.57\times {10}^{3}kg/m^3$

so option (A) is correct