Question

If force $F$, mass $M$ and time $T$ are chosen as fundamental quantities the dimensional formula for length is:

• A. $FMT$
• B. ${FM}^{-1}{T}^2$
• C. ${FL}^2{T}^{-2}$
• D. $F^{-1}{L}^{-2}{T}^{-2}$

Answer 1

The correct option is B ${FM}^{-1}{T}^2$

$M^0{L}^1{T}^0=F^{\alpha }{M}^{\beta }{T}^{\gamma }$
$M^0{L}^1{T}^0=({MLT}^{-2}{)}^{\alpha }{M}^{\beta }{T}^Y$
$M^0{L}^1{T}^0=M^{\alpha +\beta }{L}^{\alpha }{T}^{-2\alpha +y}$
$\Rightarrow \alpha +\beta =0,\alpha =1\Rightarrow \beta =-1$
$\Rightarrow -2\alpha +y=0\Rightarrow \gamma =2$
$\Rightarrow L^1=F^1{M}^{-1}{T}^2$