The $SI$ unit of energy is $J = kg \, m^{2} \, s^{-2}$; that of speed $v$ is $m \, s^{-1}$ and of acceleration $a$ is $m \, s^{-2}$. Which of the formulae for kinetic energy $(K)$ given below can you rule out on the basis of dimensional arguments ($m$ stands for the mass of the body):
$(a)$ $K = m^{2} v^{3}$
$(b)$ $K = (1/2) m v^{2}$
$(c)$ $K = m a$
$(d)$ $K = (3/16) m v^{2}$
$(e)$ $K = (1/2) m v^{2} + m a$

(A, C, E) The dimension of kinetic energy $K$ is $[M L^{2} T^{-2}]$.
According to the principle of homogeneity of dimensions,every term in an equation must have the same dimensions.
For $(a)$,the dimension is $[M^{2} (L T^{-1})^{3}] = [M^{2} L^{3} T^{-3}]$,which is incorrect.
For $(b)$,the dimension is $[M (L T^{-1})^{2}] = [M L^{2} T^{-2}]$,which is correct.
For $(c)$,the dimension is $[M (L T^{-2})] = [M L T^{-2}]$,which is incorrect.
For $(d)$,the dimension is $[M (L T^{-1})^{2}] = [M L^{2} T^{-2}]$,which is correct.
For $(e)$,the terms $m v^{2}$ and $m a$ have different dimensions ($[M L^{2} T^{-2}]$ and $[M L T^{-2}]$ respectively),so they cannot be added. This is incorrect.
Therefore,formulae $(a)$,$(c)$,and $(e)$ are ruled out based on dimensional analysis.