Units and units conversion

SI base units

SI common derived units

Unit conversion

Conversion is nothing but dividing out the unit that you don't want and multiplying the unit that you wanted.

Example 1:What is the height of a $48 \,ft$ building in $m$?.

$1\: ft=0.3048\: m$

Take the original number and multiply with $1$.

$48 \,ft=48 \,ft \times 1$

The unit that you want is $m$ and you do not want is $ft$, which you should cancel out. So, you need to bring the $ft$ to the denominator in the conversion factor.

In the conversion factor, the unit $ft$ is on the left hand side. So, divide both sides of the conversion factor by everything on the left hand side ($1 ft$):

$1\:=\dfrac{0.3048\:m}{1 ft}$

Now, replace the $1$ in the previous equation with this $1$,

$48ft = 48ft\times 1$

   $=48ft\times \dfrac{0.3048m}{1 ft}$

Canceling out $ft$,

   $=48\cancel{ft}\times \dfrac{0.3048m}{1 \cancel{ft}}$

   $=14.6304 m$

Rounding to correct number of significant figures,

$\boxed{48ft = 15 m}$

Note that when you are doing conversion, round the final result to have the same number of significant figures as the original number. Don't worry about the numbers in the conversion factor as we consider them as exact and have infinite number of significant figures.

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In the previous problem, the quantity height has only one base unit, $ft$ that we converted to $m$. If you want to convert a quantity that has more than one unit (a combination of units), then you will need more than one conversion factor but the same procedure applies.

Example 2: The speed of a car is $75 \,km/h$. What is its speed in $m/s$.

Solution:

$1\:km=1000\:m$

$1\:h=3600\:s$

Since there are two base units in the quantity, you should multiply two $1$'s with the original number and later will replace these with the conversion factors.

$75\:km/h=75\:km/h\times 1 \times 1$

Since $km$ is in the numerator, you need to have that in the denominator in the conversion factor to cancel that out. And, the $h$ is in the denominator, so you need to have that in the numerator in the conversion factor.

So, you have,

$1=\dfrac{1000\,m}{1\,km}$

$\dfrac{1\,h}{3600\,s}=1$

Replacing the $1$'s with these values, we get,

$75\:km/h=75\:km/h\times \dfrac{1000\:m}{1\:km} \times \dfrac{1\:h}{3600\:s}$

Canceling out $km$ and $h$,

$75\:km/h=75\:\cancel{km}/\cancel{h}\times \dfrac{1000\:m}{1\:\cancel{km}} \times \dfrac{1\:\cancel{h}}{3600\:s}$

Simplifying, and rounding to correct number of significant figures,

$\boxed{75\:km/h=21\: m/s}$

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There are physical quantities with units having exponents. For example, SI unit of density is $kg/m^3$. The following example will help you how to convert such quantities from one system of units to another.

Example 3: In the cgs system, unit of density is $g/cm^3$ ($g$ is grams). Density of mercury is $13.6 \, g/cm^3$. What is the density of mercury in the SI unit?

Solution:

In SI, the unit of density is $kg/m^3$. So, we need to convert $g$ to $kg$ and $cm$ to $m$ (i.e., $cm^3$ to $m^3$).

The conversion factors (from the table at the bottom of this page) are

$1 \,kg = 1000\,g$    and

$100 \,cm = 1\,m$

Since there are two units to convert, multiply the given number by two 1's. Since one of the unit has an exponent of 3, put the exponent to one of the 1's:

$13.6\,g/cm^3 = 13.6\,g/cm^3\times 1\times 1^3$

The unit $g$ is in the numerator, so to cancel that bring that to the denominator in the conversion factor. And, since the unit $cm$ is in the denominator, it should be in the numerator in the conversion factor. i.e.,

$\dfrac{1\,kg}{1000\,g}=1$    and

$\dfrac{100\,cm}{1\,m}=1$

Substituting these in place of 1's in the equation:

$13.6\,g/cm^3 = 13.6\,g/cm^3\times \dfrac{1\,kg}{1000\,g} \times \left(\dfrac{100\,cm}{1\,m}\right)^3$

Canceling, $g$ and $cm^3$.

$13.6\,g/cm^3 = 13.6\,\cancel{g}/\cancel{cm^3}\times \dfrac{1\,kg}{1000\,\cancel{g}} \times \left(\dfrac{100\,\cancel{cm}}{1\,m}\right)^3$

      $= \dfrac{13.6}{1000}\times 100^3\,kg/m^3$

      $= 1.36\times 10^4\,kg/m^3$

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Table of conversion factors