Question

The density of aluminum is $2.7gm/c{{m}^{3}}$. It’s density in $kg/{{m}^{3}}$will be:
$\begin{aligned} & a)27kg/{{m}^{3}} \\\ & b)2700kg/{{m}^{3}} \\\ & c)270kg/{{m}^{3}} \\\ & d)27000kg/{{m}^{3}} \\\ \end{aligned}$

Answer 1

In the above question we are basically asked to express the density of aluminum in the MKS system of units from the CGS system of units. First we will see how to express the grams in terms of kg and then centimeter to meter. Accordingly further we will obtain the unit of density in terms of kilogram and meter cube i.e. in SI units.

Complete step by step answer:
The MKS system of units is accepted to be the standard form of any units of the physical quantity. In this system we basically express the mass of the body in terms of kilogram, distance or the length in terms of meters and time in terms of seconds. Now in the above question we are asked to write the density of aluminum i.e. its mass in terms of kilogram and volume in terms of meters. So here are conversions that we have to make in order to write the quantity in terms of the MKS system of units.
$\begin{aligned} & 1gm={{10}^{-3}}kgm \\\ & 1cm={{10}^{-2}}m \\\ \end{aligned}$
Hence the density of aluminum in terms of $kg/{{m}^{3}}$ is equal to,
$\begin{aligned} & 2.7gm/c{{m}^{3}} \\\ & \Rightarrow \dfrac{2.7\times {{10}^{-3}}kgm}{{{({{10}^{-2}}m)}^{3}}} \\\ & \Rightarrow \dfrac{2.7\times {{10}^{-3}}}{{{10}^{-6}}}kgm/{{m}^{3}} \\\ & \therefore 2.7\times {{10}^{3}}kgm/{{m}^{3}}=2700kgm/{{m}^{3}} \\\ \end{aligned}$
Therefore we can conclude that the correct answer of the above question is option b.

Note:
It is to be noted that the density of the substance is defined as mass per unit volume. The mass is expressed in terms of kilograms and the volume is expressed as the cube of the unit length of the body. It is also to be noted that while conversion of units to some other form, it is to be noted that the power of the fundamental quantity is always to be considered.