Question

By applying dimensional analysis, convert the $10J$ into ergs.

Answer 1

Hint: As joules is the SI unit of work we have to first use the dimensional formula of work. We need to convert joules into erg which is the CGS unit of work. Now using the dimensional formula of work, convert each SI term into CGS term. Equating the SI term and CGS term with a constant multiply to each of them where the SI constant is given to us now we can calculate the converted erg term.

Complete step by step answer:
In the SI system we write work as joules while in the CGS system we represent work units as erg. The amount of work done by a system we calculate with the product of force and distance. In other words we can say that when a body moved to a certain distance by an external force is termed as work done by the object.Hence we can write mathematically as,
Work = product of Force and Displacement
$W = F \times d$
Where, force is equal to $F$ and displacement is equal to $d$.
We can write force as a product of mass and acceleration.
Hence the formula of work done will be,
$W = ma \times d$
Where, mass is equal to $m$ and acceleration is equal to $a$.
Writing work in its dimensional form we will get,
$W = Mass \times \dfrac{{Length}}{{{{\left( {Time} \right)}^2}}} \times Length$
Here acceleration of the body is length per unit time square.

Now dimensions of work will be,
$W = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]$
Here let us assume the power of mass be a, length be b and time be c.Now,
$a = 1;b = 2;c = - 2$
Now converting each SI terms to CGS we will get,
$SI\,System - MSI = 1kg,LSI = 1m,TSI = 1s$
$CGS\,System - MCGS = 1kg,LCGS = 1cm,LCGS = 1s$
Now with the help of dimensional relation we can relate the SI term with the CGS term.
Hence,
$NSI\left[ {MS{I^a}LS{I^b}TS{I^c}} \right] = NCGS\left[ {MCG{S^a}LCG{S^b}TCG{S^c}} \right]$
We know the value of $NSI = 10$ and we need to calculate $NCGS = ?$

Now rearranging the above equation we will get,
$NSI\left[ {\dfrac{{MS{I^a}}}{{MCG{S^a}}}\dfrac{{LS{I^b}}}{{LCG{S^b}}}\dfrac{{TS{I^c}}}{{TCG{S^c}}}} \right] = NCGS$
$ \Rightarrow NSI\left[ {{{\left[ {\dfrac{{MSI}}{{MCGS}}} \right]}^a}{{\left[ {\dfrac{{LSI}}{{LCGS}}} \right]}^b}{{\left[ {\dfrac{{TSI}}{{TCGS}}} \right]}^c}} \right] = NCGS$
Now putting the know value we will get,
$10\left[ {{{\left[ {\dfrac{{1kg}}{{1g}}} \right]}^1}{{\left[ {\dfrac{{1m}}{{1cm}}} \right]}^2}{{\left[ {\dfrac{{1s}}{{1s}}} \right]}^{ - 2}}} \right] = NCGS$
Now wring kg into g and m into cm we will get,
$10\left[ {{{\left[ {\dfrac{{1000g}}{{1g}}} \right]}^1}{{\left[ {\dfrac{{100cm}}{{1cm}}} \right]}^2}} \right] = NCGS$
$ \therefore 10\left[ {\left[ {\dfrac{{1000g}}{{1g}}} \right]\left[ {\dfrac{{10000c{m^2}}}{{1c{m^2}}}} \right]} \right] = NCGS$

Hence from the above calculation we can conclude that $10J = {10^8}erg$.

Note: Dimensional analysis is the analysis of the relationships between different physical quantities or we can say here the physical quantities are expressed in terms of their fundamental dimensions by identifying their base quantities and units of measure and tracking these dimensions as calculations or comparisons are performed.