Question

A piece of stone of mass $15.1g$ is first immersed in a liquid and it weights $10.9gf$. Then on immersing the piece of stone in water, it weights $9.7gf$. Calculate :

A. The weight of the piece of stone in the air.

B. The volume of the piece of stone.

C. The relative density of stone.

D. The relative density of the liquid.

Answer 1

Given that,

Weight of stone in air = $15.1g$

Weight in liquid = $10.9gf$

Weight in water = $9.7gf$

Now, solve all terms

(A). The weight of the piece of stone in air

$W_{air}=15.1g$

(B). the volume of the piece of stone is

$V=\frac{W_{air}-W_{water}}{\rho }$

$V=\frac{15.1-9.7}{1}$

$V=5.4c{m}^3$

(C). the relative density of stone is

$R.D_{stone}=\frac{15.1}{15.1-9.7}$

$R.D_{stone}=\frac{15.1}{5.4}$

$R.D_{stone}=2.8$

(D). the relative density of the liquid is

$R.D_{liquid}=\frac{W_{air}-W_{liquid}}{W_{air}-W_{water}}$

$R.D_{liquid}=\frac{15.1-10.9}{15.1-9.7}$

$R.D_{liquid}=\frac{4.2}{5.4}$

$R.D_{liquid}=0.78$

Hence, the weight of the piece of stone in air will be $15.1g$.

The volume of the piece of stone will be $5.4c{m}^3$.

The relative density of stone and the relative density of the liquid will be $2.8\&0.8$ respectively.