Question

to check whether the equation Impusle=F×(∆t)^2 is dimensionally correct?​

Answer 1

$\textsf{\underline {\Large{Dimensional Correctness}}}$ :

$\textsf{\underline {Equation}}$ : $\boxed{\mathsf{\vec{J} \:=\:{\vec{F}{\times{\Delta{t}^{2}}}}}}$

For dimensional correctness, we follow the $\textsf{\underline {Principle of Homogeneity}}$ in which the given will be dimensionally correct if there will be same dimensional formula in both the sides (L. H. S. &. R. H. S.)

Dimensional Formula of Impulse = $\mathsf{[ML{T} ^{-1}}]$

Dimensional Formula of Force = $\mathsf{[ML{T} ^{-2}}]$

Now, L. H. S. = $\mathsf{[ML{T} ^{-1}}]$

R. H. S. = $\mathsf{[ML{T} ^{-1}}]$ × $\mathsf{{[T]}^{2}}$

R.H. S. = $\mathsf{[ML{T} ^{-1\:+\:2}}]$

R.H.S. = $\mathsf{[ML{T} ^{1}}]$

R.H.S. = $\mathsf{[MLT}]$

➡️ Hence, $\mathsf{[ML{T} ^{-1}}]$ ≠ $\mathsf{[MLT}]$

⚫L. H. S. ≠ R. H. S.

⚫That means, the given equation is dimensionally incorrect.