Question

If momentum [P], area [A] and time [T] are taken to be fundamental quantities, then energy has the dimensional formula
A. $\left[ {{P}^{1}}{{A}^{-1}}{{T}^{1}} \right]$
B. $\left[ {{P}^{2}}{{A}^{1}}{{T}^{1}} \right]$
C.$\left[ {{P}^{1}}{{A}^{-\dfrac{1}{2}}}{{T}^{1}} \right]$
D.$\left[ {{P}^{1}}{{A}^{\dfrac{1}{2}}}{{T}^{-1}} \right]$

Answer 1

As a first step, you can recall the dimensions of the individual quantities in the given question. Then you could write a proportionality relation of energy with momentum, area and time, all three raised to different powers (say, α, β, γ). Now substitute the dimensions of P, A and T. Now you could find the powers at which each quantity is raised using the principle of homogeneity of dimensions and hence the answer.

Complete answer:
We can define the dimensions of a physical quantity as the powers or exponents to which the fundamental/ base quantities are raised to represent that particular quantity. And we normally denote them by square brackets ‘[]’. In the given question we are asked to express energy in terms of momentum, area and time considering them as fundamental quantities.
Let us first recall the dimensions of energy, momentum, area and time in terms of actual fundamental quantities (mass, length, time, etc…).
Dimensional formulae for energy,
We have, $E=m{{c}^{2}}$
Writing this equation in terms of fundamental quantities raised to some power, that is, taking dimensions on both sides, we get,
$\left[ E \right]=\left[ M \right]{{\left[ L{{T}^{-1}} \right]}^{2}}=\left[ M{{L}^{2}}{{T}^{-2}} \right]$ ……………………….. (1)
Dimensional formulae for momentum,
We have, $P=mv$
Taking dimension on both sides,
$\left[ P \right]=\left[ M \right]\left[ L{{T}^{-1}} \right]=\left[ ML{{T}^{-1}} \right]$ ……………………….. (2)
Dimensional formulae for area,
$\left[ A \right]={{\left[ L \right]}^{2}}$ ………………………… (3)
Dimensional formulae for time,
$\left[ T \right]=\left[ T \right]$ ………………………. (4)
Let us write the dependence of energy E on the given quantities given as fundamental quantities- momentum, area and time as,
$E\propto {{P}^{\alpha }}{{A}^{\beta }}{{T}^{\gamma }}$
$E=k{{P}^{\alpha }}{{A}^{\beta }}{{T}^{\gamma }}$ ………………………. (5)
Where, k is the dimensionless constant of proportionality.
Let us now take dimensions on both sides of equation (5).
$\left[ E \right]={{\left[ P \right]}^{\alpha }}{{\left[ A \right]}^{\beta }}{{\left[ T \right]}^{\gamma }}$ ………………………….. (6)
Substituting (1), (2), (3), (4) in (6), we get,
$\left[ M{{L}^{2}}{{T}^{-2}} \right]={{\left[ ML{{T}^{-1}} \right]}^{\alpha }}{{\left[ {{L}^{2}} \right]}^{\beta }}{{\left[ T \right]}^{\gamma }}$
$\left[ M \right]{{\left[ L \right]}^{2}}{{\left[ T \right]}^{-2}}={{\left[ M \right]}^{\alpha }}{{\left[ L \right]}^{\alpha +2\beta }}{{\left[ T \right]}^{\gamma -\alpha }}$ ………………………….. (7)
By the principle of homogeneity of dimensions, the dimensions on both sides of any valid equation should be the same. So, from equation (7) we have,
$\alpha =1$ ………………….. (8)
$\alpha +2\beta =2$…………………….. (9)
$\gamma -\alpha =-2$ ……………………….. (10)
Substituting (8) in (9), we get,
$1+2\beta =2$
$\beta =\dfrac{1}{2}$ …………………… (11)
Substituting (8) in (10), we get,
$\gamma -1=-2$
$\gamma =-1$ ……………………….. (12)
Now, we can substitute the values of α, β, γ from (8), (11) and (12) in (6).
$\left[ E \right]={{\left[ P \right]}^{1}}{{\left[ A \right]}^{\dfrac{1}{2}}}{{\left[ T \right]}^{-1}}$
Therefore, the dimension of energy in terms of momentum, area and time as fundamental quantities is given by,
$\left[ E \right]=\left[ {{P}^{1}}{{A}^{\dfrac{1}{2}}}{{T}^{-1}} \right]$

Hence, the answer to the question is option D.

Note:
Though the dimensional analysis is useful in deducing relations among the interdependent quantities, dimensionless constants cannot be obtained by this method. So, by this method we can only test the dimensional validity not the exact relationship between quantities. Also, there is no distinction between the quantities with the same dimensions.