Question

A solid weighs $32gf$ in air and $28.8gf$ in water. The Relative density of solid is $X$. Find ($X-4$).

• A. $14$
• B. $10$
• C. $6$
• D. $5$

Answer 1

The correct option is C $6$

Let us consider that $D_w$ and $D$ be the density of water and solid respectively.

The actual weight of the solid is $32gf$.

Apparent weight of the body = Actual weight - Buoyant force ($V\times D_w\times g$).

Let $V$ be the volume of the solid.

Density of water $D_w=1g{cm}^3$.

In this case, $28.8=mg-V\times D_w\times g$

That is, $28.8=32-V\times 1$

Therefore, $V=3.2{cm}^3$.

The density of the body is given as mass/Volume, that is $32/3.2=10g{cm}^3$.

The relative density of a solid is given as the ratio of the density of a substance to the density of a standard substance under specified conditions.

That is, $Relativedensity=\frac{Densityofsolid}{Densityofwater}=\frac{10g{cm}^3}{1g{cm}^3}=10$

Hence, X-4 = 10-4 = 6.