
Aluminium has a density of $2700\;kg{m^{ - 3}}$. What is the density in $gc{m^{ - 3}}$?
Answer
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Hint: In the question we need to convert the given unit of density which is $kg{m^{ - 3}}$ to $gc{m^{ - 3}}$ and for doing this conversion, we first need to write the conversion as a fraction and then multiply it out with the proper conversion factor and at last, cancel units which are present at the top as well as bottom of the fraction.
Complete answer:
As per question, it is given that the density of aluminium is $2700\;kg{m^{ - 3}}$. The density of aluminium can be represented in $gc{m^{ - 3}}$ by following several steps which are as follows:
Step-1: Converting units into fraction:
We have to represent the given units into fraction as follows-
$\rho = 2700\dfrac{{kg}}{{{m^3}}}$
Step-2: Determining conversion factors (that equals one) for each quantity in the given unit:
We know that conversion factors for kg and m are as follows-
$1kg = 1000g$
$ \Rightarrow \dfrac{{1000g}}{{1kg}} = 1$
$1m = 100cm$
$ \Rightarrow \dfrac{{1m}}{{100cm}} = 1$
Step-3: Multiply the given value with proper conversion factors without cancelling the units:
We have to multiply the conversion factor of metre three times because the power of metre in the initial unit is given as 3.
$\rho = 2700\dfrac{{kg}}{{{m^3}}} \times \dfrac{{1000g}}{{1kg}} \times \dfrac{{1m}}{{100cm}} \times \dfrac{{1m}}{{100cm}} \times \dfrac{{1m}}{{100cm}}$
Step-4: Cancel out the units which are common in both numerator as well as denominator in order to get the final answer.
$\rho = 2.7\dfrac{g}{{c{m^3}}}$
Hence, the density of aluminium in $gc{m^{ - 3}}$ is $2.7gc{m^{ - 3}}$.
Note:
It is important to note that while equating conversion factors with one, the unit in numerator and denominator should be decided in such a way that the undesired unit or the unit we need to change must cancel out. For example, in this question, if we write the conversion factor of kg upside down, i.e., $\dfrac{{1kg}}{{1000g}} = 1$, then the unit kg will never cancel out and can lead to calculation mistakes.
Complete answer:
As per question, it is given that the density of aluminium is $2700\;kg{m^{ - 3}}$. The density of aluminium can be represented in $gc{m^{ - 3}}$ by following several steps which are as follows:
Step-1: Converting units into fraction:
We have to represent the given units into fraction as follows-
$\rho = 2700\dfrac{{kg}}{{{m^3}}}$
Step-2: Determining conversion factors (that equals one) for each quantity in the given unit:
We know that conversion factors for kg and m are as follows-
$1kg = 1000g$
$ \Rightarrow \dfrac{{1000g}}{{1kg}} = 1$
$1m = 100cm$
$ \Rightarrow \dfrac{{1m}}{{100cm}} = 1$
Step-3: Multiply the given value with proper conversion factors without cancelling the units:
We have to multiply the conversion factor of metre three times because the power of metre in the initial unit is given as 3.
$\rho = 2700\dfrac{{kg}}{{{m^3}}} \times \dfrac{{1000g}}{{1kg}} \times \dfrac{{1m}}{{100cm}} \times \dfrac{{1m}}{{100cm}} \times \dfrac{{1m}}{{100cm}}$
Step-4: Cancel out the units which are common in both numerator as well as denominator in order to get the final answer.
$\rho = 2.7\dfrac{g}{{c{m^3}}}$
Hence, the density of aluminium in $gc{m^{ - 3}}$ is $2.7gc{m^{ - 3}}$.
Note:
It is important to note that while equating conversion factors with one, the unit in numerator and denominator should be decided in such a way that the undesired unit or the unit we need to change must cancel out. For example, in this question, if we write the conversion factor of kg upside down, i.e., $\dfrac{{1kg}}{{1000g}} = 1$, then the unit kg will never cancel out and can lead to calculation mistakes.
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