Question

The relative density of silver is $10.5$, if the density of water is $1000\mathrm{kg}{\mathrm{m}}^{-3}$, then density of silver will be :

• A. $10.5\mathrm{kg}{\mathrm{m}}^{-3}$
• B. $1050\mathrm{kg}{\mathrm{m}}^{-3}$
• C. $10.5\mathrm{kg}{\mathrm{m}}^3$
• D. $10500\mathrm{kg}{m}^{-3}$

Answer 1

The correct option is D

$10500\mathrm{kg}{m}^{-3}$

Step 1: Given data

Relative density of silver = 10.5

Density of water = $1000Kg/m^3$

Step 2: Finding the density of silver

Relative density, or specific gravity, is the ratio of the density (mass of a unit volume) of a substance to the density of water.

$R\mathrm{elative}\mathrm{density}\mathrm{of}\mathrm{silver}\left(RD\right)=\frac{\mathrm{density}\mathrm{of}\mathrm{silver}\left({\mathrm{d}}_{\mathrm{silver}}\right)}{\mathrm{density}\mathrm{of}\mathrm{water}\left({\mathrm{d}}_{water}\right)}$

${\mathrm{d}}_{\mathrm{silver}}=RD\times {\mathrm{d}}_{\mathrm{water}}$

$=10.5\times 1000kg{\mathrm{m}}^{-3}$

$=10500kg{\mathrm{m}}^{-3}$

Hence, the correct option is (D).