Question

If force $F$, acceleration $a$, and time $T$ are taken as the fundamental physical quantities, the dimensions of length on this system of units are

• A. $FA{T}^2$
• B. $FAT$
• C. $FT$
• D. $A{T}^2$

Answer 1

The correct option is D $A{T}^2$

Force $=F=\left[{MLT}^{-2}\right]$

acceleration $=A=\left[{LT}^{-2}\right]$

time $=T=\left[T\right]$

length $=L$

$L=F^x{A}^y{T}^z$

$\left[L\right]={\left[{MLT}^{-2}\right]}^x{\left[{LT}^{-2}\right]}^y{\left[T\right]}^z$

$\left[M^0{LT}^0\right]=\left[M^x{L}^{x+y}{T}^{-2x-2y+z}\right]$

Comparing both sides,

$x=0$ $x+y=1$

$y=1$

$-2x-2y+z=0$

$0-2\times 1+z=0$

$z=2$

Length $=F^0{A}^1{T}^2$

$L=\left[F^0{AT}^2\right]=\left[{AT}^2\right]$