Question

A $0.50N$ metal sinker appears (as measured using a spring scale) to have a weight of $0.45N$ when submerged in water.Find out the relative density of the metal .

• A. $6$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (D)

$10$

Answer is D.

Let us consider that Dw and D be the density of water and solid respectively.
The actual weight of the solid is 0.50 N
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 $D_w=1g/c{m}^3$.
In this case, $0.45=mg-V\times D_w\times g$
That is, $0.45=0.50-V\times 1$
Therefore, V = $0.05{cm}^3$.
The density of the body is given as Mass/Volume, that is, $0.50/0.05=10g{cm}^3$.
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, the relative density of the solid is 10.