Q4E - University Physics with Modern Physics

Question

The density of gold is \(19.3\,{{\rm{g}} \mathord{\left/

 {\vphantom {{\rm{g}} {{\rm{c}}{{\rm{m}}^{\rm{3}}}}}} \right.

 \kern-\nulldelimiterspace} {{\rm{c}}{{\rm{m}}^{\rm{3}}}}}\). What is this value in kilograms per cubic meter?

Step-by-Step Solution

Answer

The density of gold is \(1.93 \times {10^4}\,{{{\rm{kg}}} \mathord{\left/

 {\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.

 \kern-\nulldelimiterspace} {{{\rm{m}}^{\rm{3}}}}}\)

Step 1: Density and its unit

Density is the ratio of the mass of the body to its volume. Here hence if we take mass as kg and volume as m3, than the SI unit of density becomes $kg/m{g}^3$

Density of gold is, $pkg/m^3$

Given data 

We know that the conversion of unit is as per below:

$$\begin{gathered} 1kg=1000g \\ 1m= 1000cm \end{gathered}$$

Step 2: Determination of density in SI unit

Density of the gold is given $19.3g/c{m}^3$

$$\begin{gathered} 1kg=1000g \\ 1m\equiv 1000cm \end{gathered}$$

Then the density of gold is,

p$p=19.3g/c{m}^3\times \frac{1kg}{1000g}\times {\left(\frac{100cm}{1m}\right)}^3$

$1.93\times {10}^{4}kg/m^3$

Hence from the above calculation we can say $1.93\times {10}^{4}kg/m^3$is the value of density in kilograms per cubic meter.