Question

If momentum $P$, area $A$ and time$T$ are taken to be the fundamental quantities then the dimensional formula for energy is

• A. $\left[P^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}A{T}^{-1}\right]$
• B. $\left[P{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{T}^{-1}\right]$
• C. $\left[PAT\right]$
• D. $\left[PA{T}^2\right]$

Answer 1

The correct option is B

$\left[P{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{T}^{-1}\right]$

Step 1: Given data

Given, momentum, area and time are the fundamental quantities.

We have to find the dimensional formula of energy.

Step 2: Formula and solution

Dimension of energy,

$$\begin{gathered} E=\frac{1}{2}m{v}^2 \\ dimofenergy=\left[M{L}^2{T}^{-2}\right] \\ dimofmomentum=\left[ML{T}^{-1}\right] \\ dimofarea=\left[L^2\right] \\ dimoftime=\left[T\right] \end{gathered}$$

Considering $X,Y,Z$ to be the power of $P,A,T$ we get

$$\begin{gathered} \left[M{L}^2{T}^{-2}\right]=\left[P^x{A}^Y{T}^Z\right] \\ \Rightarrow M{L}^2{T}^{-2}={\left(ML{T}^{-1}\right)}^X{\left(L^2\right)}^Y{\left(T\right)}^Z \\ \Rightarrow M{L}^2{T}^{-2}=M^X{L}^{X+2Y}{T}^{-X+Z} \end{gathered}$$

Now, equating the values,

$$\begin{gathered} \Rightarrow X=1 \\ \Rightarrow X+2Y=2 \\ 1+2Y=2 \\ Y=\frac{1}{2} \\ \Rightarrow -X+Z=-2 \\ -1+Z=-2 \\ Z=-1 \end{gathered}$$

Therefore, the dimensional formula is $\left[P{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{T}^{-1}\right]$

Hence, option B is correct.