The SI unit of energy is `J=kg m^(2)s^(-2)`, that of speed v is `ms^(-1)` and of acceleration a is `ms^(-2)`. What 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^(2)`
(b) `K = (1/2)mv^(2)`
`("c") K= ma`
(d) `K = (3/16)mv^(2)`
(e) `K=(1/2)mv^(2)+ma`

Explanation

Only quantities with the same physical dimensions can be added or subtracted. Every correct formula or equation must have the same dimension on both sides of the equation.
LHS dimension of kinetic energy `K=[M L^(2)T^(-2)]`
RHS dimension of `(a)==M^(2)(L^(2)T^(-1))^(3)`
`=M^(2)L^(3)T^(-3)`
RHS dimension of `(b)=(1)/(2)mv^(2)`
`=(1)/(2)`= constant
= dimensionless
`=M^(1)(L^(1)T^(-1))^(2)`
`=M^(1)L^(2)T^(-2)`
RHS dimension of `(c)=ma`
`=M^(1)L^(1)T^(-2)`
RHS dimension of (d) `=(3)/(16) mv^(2)`
`(3)/(16)`= constant
=dimensionless
`=mv^(2)`
`=M^(1)(L^(2)T^(-1))^(2)`
`=M^(1)L^(2)T^(-2)`
RHS dimensio of (e) first term `=(1)/(2) mv^(2)`
`=M^(1)L^(1)T^(-2)`
second term `=M^(1)L^(1)T^(-2)`
Here both term of (e) have difference.
Here, (a), (c), (d) have different dimension and not that of kinetic energy. In (b) and (d) dimension of LHS and RHS are equal but by definition of kinetic energy (d) is not correct.
Hence, in all respect.