Question

A solid weighs $120gf$ in air and $105gf$ when it is completely immersed in water. Calculate the relative density of solid.

• A. $1/8$
• B. $8$
• C. $7/8$
• D. $8/7$

Answer 1

The correct option is B $8$

The solid apparently weighs less in water when compared to air because of the upward buoyant force acting on it when it is submerged in water.
According to Archimedes principle, the buoyant force acting on an object is equal to the weight of the fluid displaced by it.
Therefore, the apparent decrease in the weight of an object is equal to the weight of the fluid displaced by it.
Weight of displaced water = Apparent decrease in weight = $120-105=15$ $gf$
Mass of displaced water = 15g
Volume of solid = volume of water displaced = $\frac{mass}{density}=\frac{15g}{1g/{cm}^3}=15{cm}^3$
Actual weight of solid = weight of solid in air =120$gf$ (since buoyant force due to air is negligible)
Mass of solid = 120g
Density of solid = $\frac{Mass}{Volume}=\frac{120}{15}=8g/{cm}^3$
Relative density of solid = $\frac{{Density}_{solid}}{{Density}_{water}}=\frac{8g/{cm}^3}{1g/{cm}^3}=8$