Questions & Answers

Question

Answers

$\begin{align}

& \text{A}\text{. 1}\times \text{1}{{\text{0}}^{3}}kg{{m}^{-3}} \\

& \text{B}\text{. 10}\text{.8}\times \text{1}{{\text{0}}^{3}}kg{{m}^{-3}} \\

& \text{C}\text{. 108}\times \text{1}{{\text{0}}^{3}}kg{{m}^{-3}} \\

& \text{D}\text{. 54}\times \text{1}{{\text{0}}^{3}}kg{{m}^{-3}} \\

\end{align}$

Answer
Verified

Density of a body or material can be defined as the ratio of the mass and the volume occupied by the body. The relative density of a material can be defined as the density of a material with respect to the density of a reference material. The SI unit of density is $kg{{m}^{-3}}$ while the relative density us a dimensionless quantity.

In the question, the relative density of silver is given as,

${{\rho }_{sr}}=10.8$

The relative density of the silver is given as the ratio of the density of the silver to the density of the water. So, we can write that,

$\text{relative density of silver = }\dfrac{\text{density of silver}}{\text{density of water}}$

Now, the density of the water is given as, ${{\rho }_{w}}=1\times {{10}^{3}}kg{{m}^{-3}}$

Now, putting the given values on the above equation, we get that,

$\begin{align}

& {{\rho }_{sr}}=\dfrac{{{\rho }_{s}}}{{{\rho }_{w}}} \\

& {{\rho }_{s}}={{\rho }_{sr}}\times {{\rho }_{w}} \\

& {{\rho }_{s}}=10.8\times 1\times {{10}^{3}}kg{{m}^{-3}} \\

& {{\rho }_{s}}=10.8\times {{10}^{3}}kg{{m}^{-3}} \\

\end{align}$

So, the density of the silver is $10.8\times {{10}^{3}}kg{{m}^{-3}}$.

The reference material selected to find the relative density can be anything. For solid and liquid, to find the relative density the reference material is water. For gas, to find the relative density, the reference material is air.