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(The question has multiple correct options)

A. $K=m^2 v^3$

B. $K= (1/2) mv^2$

C. $K=ma$

D. $K=(3/16) m v^2$

E. $K= (1/2) mv^2 + ma$

Answer
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Given,

$J= kg m^2 s^{-2}$

The dimensions for this expression can be written as $[ML^2T^{-2}]$. As we simply translated the respective units into dimensions like kg became M, m (not mass here) corresponds to L and s is a unit of time so T was placed.

We analyze all the options one by one dimensionally;

(A) $K=m^2 v^3$

m is mass, v is velocity so, the dimensions are $[M^2L^3T^{-3}]$.

This clearly does not match with $[ML^2T^{-2}]$.

(B) $K= (1/2) mv^2$

This is our familiar kinetic energy expression, so it is obviously correct as dimensions are $[ML^2T^{-2}]$.

(C) $K=ma$

We are aware that ma is used for force. The dimensions will be $[MLT^{-2}]$. So this can be discarded too.

(D) $K=(3/16) m v^2$

This might not look as familiar as (B) but this is a dimensionally correct kinetic energy expression. Only constant in the front is 3/16 here in place of 1/2. This does not affect dimensional analysis.

(E) $K= (1/2) mv^2 + ma$

An important point to be noted from this expression is that the dimensions of all the terms on the LHS and RHS should match. The second term in the RHS is ma, which will give dimensions of force i.e., $[MLT^{-2}]$. Therefore making the entire expression dimensionally incorrect.

Therefore, the correct answers to be chosen here are (A), (C) and (E) as all these can be ruled out based on dimensional arguments.

Read the question carefully. It is specifically asked to choose the wrong expression. One can tick the correct expression instead, when in a hurry. Also the (3/16) factor does not affect the dimensions so that should be kept in mind.