Question

Suppose speed of light $c$, force $F$ and kinetic energy $K$ are taken as the fundamental units, then the dimensional formula for mass will be
(A) $K{C^{ - 2}}$
(B) $K{F^{ - 2}}$
(C) $C{K^{ - 2}}$
(D) $F{C^{ - 2}}$

Answer 1

Hint We solve this problem by using the method of dimensional analysis. We find the dimensional formula of the qualities mentioned in the question. We equate the product of these dimensional formulas to the dimensional formula of mass. Solving the equation and finding the powers we get the dimensional formula of mass in terms of speed of light, force, and kinetic energy.

Complete step by step answer:
Speed of light has units of \[m/s\](meters per second)
Its dimensional formula is
\[C = [L{T^{ - 1}}]\]
The units of force are newton or \[kgm/{s^2}\] (kg meter per second square)
The dimensional formula of force is
\[F = [{M^1}{L^1}{T^{ - 2}}]\]
Unit of kinetic energy is $kg{m^2}/{s^2}$(kg meter square per second square or joules)
The dimensional formula of kinetic energy is
$K = [{M^1}{L^2}{T^{ - 2}}]$
Here,
Mass, Length, time are represented by$M,L,T$ respectively
The mass has a dimensional formula of
$M = [{M^1}]$
Equating the dimensional formula of mass with the product of other quantities
\[M = {C^\alpha }{F^\beta }{K^\gamma }\].....(1)
\[[{M^1}] = {[L{T^{ - 1}}]^\alpha }{[{M^1}{L^1}{T^{ - 2}}]^\beta }{[{M^1}{L^2}{T^{ - 2}}]^\gamma }\]
\[\Rightarrow [{M^1}] = [{M^{\beta + \gamma }}{L^{\alpha + \beta + 2\gamma }}{T^{ - \alpha - 2\beta - 2\gamma }}]\]
Solving for the powers
\[ \;\beta + \gamma = 1\]
\[\alpha + \beta + 2\gamma = 0\]
\[\alpha + 2\beta + 2\gamma = 0\]

Solving the above three equations,
$\beta = 0$
$\gamma = 1$
$\alpha = - 2$

Substituting the powers back in equation (1)
\[[M] = [{C^{ - 2}}{K^1}]\]

Hence option (A) $K{C^{ - 2}}$is the correct answer.

Additional information An equation, which gives the relation between fundamental units and derived units in terms of dimensions is called dimensional formula. In terms of mechanics length, mass, time are taken as the fundamental units.

Note Dimensional analysis can be very useful to solve any problem. Using dimensional analysis, we can find the units of any quantity. Dimensional analysis can be very handy even for cross-checking the final answer units. This way we can be sure that the answer we found is correct.