Question

If pressure $P$, area $A$ and length $L$ are taken to be fundamental quantities in a system of units, then energy has dimensional formula.

• A. $P^1{A}^1{L}^1$
• B. $P^1{A}^{3/2}{L}^0$
• C. $P^1{A}^0{L}^3$
• D. Cannot take $P,A$ and $L$ as fundamental quantities

Answer 1

Answer: (B)

$P^1{A}^{3/2}{L}^0$

We know that

$P=\frac{P}{A}=\frac{ML{T}^{-2}}{L^2}$

$=M{L}^{-1}{T}^{-2}$

$A=L^2$

$L=L$

Let $E=P^x{A}^y{L}^z$

$E=M{L}^2{T}^{-2}$

$=\left(M{L}^{-1}{T}^{-2}{)}^x\right(L^2{)}^y(L{)}^2$

$=M^x{L}^{-x+2y+z}{T}^{-2x}$

$x=1,-2x=-2$

$x=1$

$-x+2y+z=2$

$z=0,2y=2+1$

$y=\frac{3}{2}$

$\therefore E=P^1{A}^{3/2}{L}^0$.