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. $\left[p^1{A}^{-1/2}{t}^1\right]$
• D. $\left[p^1{A}^{1/2}{t}^{-1}\right]$

Answer 1

Answer: (D)

$\left[p^1{A}^{1/2}{t}^{-1}\right]$

Let, energy $E=k{p}^a{A}^b{t}^c$ ...(i)
where k is a dimensionless constant of proportionality
Equating dimensions on both sides of (i), we get
$\left[M{L}^2{T}^{-2}\right]=\left[ML{T}^{-1}{]}^a\right[M^0{L}^2{T}^0{]}^b[M^0{L}^{0}T{]}^c$
$=\left[M^a{L}^{a+2b}{T}^{-a+c}\right]$
Applying the principle of homogeneity of dimensions,
we get
$a=1$ ...(ii)
$a+2b=2$ ...(iii)
$-a+c=-2$ ...(iv)
On solving eqs. (ii), (iii) and (iv), we get
$a=1,b=\frac{1}{2},c=-1$
$\therefore \left[E\right]=\left[p^1{A}^{1/2}{t}^{-1}\right]$