**Question:**

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

$\left[\mathrm{P}^{2} \mathrm{AT}^{-2}\right]$

$\left[\mathrm{PA}^{-1} \mathrm{~T}^{-2}\right]$

$\left[\mathrm{PA}^{1 / 2} \mathrm{~T}^{-1}\right]$

$\left[\mathrm{P}^{\mathrm{1} / 2} \mathrm{AT}^{-1}\right]$

Question of from chapter.

JEE Main Previous Year 2020

Correct Option: 3

**Solution:**

Energy, $E \propto A^{a} T^{b} P^{c}$

or, $\quad E=k A^{a} T^{b} P^{c}$

where $k$ is a dimensionless constant and $a, b$ and $c$ are the exponents.

Dimension of momentum, $P=M^{1} L^{1} T^{-1}$

Dimension of area, $A=L^{2}$

Dimension of time, $T=T^{1}$

Putting these value in equation (i), we get

$M^{1} L^{2} T^{-2}=M^{c} L^{2 a+c} T^{b-c}$

by comparison

$\begin{aligned}

&c=1 \\

&2 a+c=2 \\

&b-c=-2 \\

&c=1, a=1 / 2, b=-1 \\

&\therefore E=A^{1 / 2} T^{-1} P^{1}

\end{aligned}$

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