Question

Class 11th
check the dimentional of
lhs=dimentional of rhs
If,
1) 1/2mv^2=mg^2
2) s=ut+1/2at^2

Answer 1

hello mate

in ur first question lhs is not equal to rhs

see ML^2T^-2 = ML^2T^-4

2- IN and question it's equal to dimension of length that is L

Answer 2

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

1. $\mathsf{\dfrac{1}{2}m{v}^{2}\:=\:m{g} ^{2}}$

L.H.S. = Dimension of $\mathsf{m{v} ^{2}}$

L.H.S. = $\mathsf{[M][{LT} ^{-1}]}$

R. H. S. = Dimension of $\mathsf{m{g} ^{2}}$

R.H.S. = $\mathsf{[M][L{T}^{-2}}]$

Hence, $\mathsf{[M][{LT} ^{-1}]}$ ≠ $\mathsf{[M][L{T}^{-2}}]$

So, First equation is dimensionally incorrect and doesn't follow Principle of Homogeneity.

( 2 ). $\mathsf{s\:=\:ut\:+\:{\dfrac{1}{2}}a{t}^{2}}$

L. H. S. = Dimension of $\textsf{s}$

L. H. S. = $\mathsf{[L]}$

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

We know that, by the Principle of Homogeneity of dimensions, that two physical quantities can be added, subtracted or compared, when they are of same kind.

So,

Dimension of $\textsf{ut}$ = Dimension of $\mathsf{a{t}^{2}}$

$\mathsf{[L{T} ^{-1}][T]}$ = $\mathsf{[L{T}^{-2}][{T} ^{2}]}$

$\mathsf{[L]}$ = $\mathsf{[ L]}$

Hence, L. H. S. = R. H. S.

So, the equation is dimensionally correct.

$\textsf{\large{\underline {Must\:know}}}$ :

$\textsf{\underline {Principle of homogeneity}}$ :

It states that a given equation will be dimensionally correct when the dimensions of each physical quantity on both sides will be equal.

Also, only "like" or same kind of physical quantities can be added, subtracted or compared.

$\textsf{\underline{Example}}$ - Momentum can be added, subtracted or compared with a physical quantity having same dimensions as of Momentum but not to any different one like "Force".