Question

Which of the following is dimensionally incorrect ?

• A. $u=v-at$
• B. $s-ut=\frac{1}{2}a{t}^2$
• C. $u^2=2a(gt-1)$
• D. $v^2-u^2=2as$

Answer 1

Answer: (C)

$u^2=2a(gt-1)$

A> We know acceleration a=rate of change of velocity

$a=\frac{v-u}{t}$ where v=final velocity and u = initial velocity and t= time

Thus we can rewrite the above equation as,

$u=v-at$

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

$\left[L{T}^{-1}\right]=\left[L{T}^{-1}\right]$ (Dimensionally correct)

B> From equation of motion we get displacement

$s=ut+\frac{1}{2}a{t}^2$

$s-ut=\frac{1}{2}a{t}^2$

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

Here also, LHS =RHS

C>The equation $u^2=2a(gt-1)$

Substituting the dimensions:

$[L{T}^{-1}{]}^2=[L{T}^{-2}\left]\right(\left[L{T}^{-2}\right]\left[T\right])$

$\left[L^2{T}^{-2}\right]=\left[L^2{T}^{-3}\right]$

LHS is not equal to RHS.

D> From equation of motion,

$v^2=u^2+2as$

$v^2-u^2=2as$

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

So, LHS=RHS

Whereas the dimensions of the equation are not the same on the LHS and RHS.

Thus option C is incorrect