Question

How much \[1\,{\text{gm}}/{\text{cc}}\] \[ = \]______ \[{\text{kg}}/{{\text{m}}^3}\]

Answer 1

Hint:First of all, we will express the given density in a fraction. After that we will convert the two units into an S.I system of equivalent units. We will manipulate and simplify accordingly to obtain the desired result.

Complete step by step answer:
We know that dimensions of density are \[M{L^{ - 3}}\] .
So, \[1\,{\text{kg}} = 1000\,{\text{g}}\] and \[1\,{\text{m}} = 100\,{\text{cm}}\]
Thus,we have,
$1\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}} \\
\Rightarrow \dfrac{{1\,{\text{g}}}}{{1\,{\text{c}}{{\text{m}}^3}}} \\
\Rightarrow\dfrac{{{{10}^{ - 3}}\,{\text{kg}}}}{{1 \times {{\left( {{{10}^{ - 2}}} \right)}^3}\,{{\text{m}}^3}}} \\
\Rightarrow \dfrac{{{{10}^{ - 3}}\,{\text{kg}}}}{{{{10}^{ - 6}}\,{{\text{m}}^3}}} \\$
\[ \therefore {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}}\]

Hence, the required answer is \[1\,{\text{gm}}/{\text{cc}} = 1 \times {10^3}\,{\text{kg}}/{{\text{m}}^3}\] .

Additional information:
Density: The density of a material (more specifically, the volumetric mass density, also known as real mass) is its mass per unit volume. The symbol most commonly used for density is the symbol (the lower-case Greek letter rho), although it is also possible to use the Latin letter D. Density is defined mathematically as mass divided by volume \[\rho = \dfrac{m}{V}\] . Where, \[\rho \] is the density, \[m\] is the mass and \[V\] is the volume.

Note:Remember that when finding out if anything is going to float in water, density is critical and it can also be useful for measuring the mass of a certain volume of a substance. A material’s density varies with temperature and pressure. For solids and liquids, this difference is usually small but much larger for gases. Increasing the weight on an object reduces the object’s volume and thereby raises its density.For interchangeability and global exchange, dimensional measurement is of basic importance. It is how we make sure things are going to work together. It would not be possible to globalize the industry without global length specifications as the basis for standardized parts.