Question

If acceleration $\left( A \right)$, mass $\left( M \right)$, velocity $\left( V \right)$, time $\left( T \right)$ are the fundamental units, then the dimensions of height will be
$A)\text{ }V{{T}^{-1}}{{M}^{2}}$
$B)\text{ }AM{{T}^{-1}}$
$C)\text{ }{{\text{A}}^{-1}}{{M}^{3}}{{V}^{-1}}$
$D)\text{ }A{{T}^{2}}$

Answer 1

This problem can be solved by writing the dimensions of height (length) in the fundamental units and also writing the dimensions of the given new fundamental quantities in the fundamental units while assigning them powers. Upon comparing the dimensions, we will get the values of the exponents and hence, get the required answer.

Complete step by step answer:
We will write the dimensions of height (that is length) according to the original fundamental units and then compare it to the new fundamental quantities' dimension after assigning them some random powers which we will then find out.
Hence, let us proceed to do that.
The dimension of height is $\left[ {{M}^{0}}L{{T}^{0}} \right]$.
The dimension of acceleration $\left( A \right)$ is $\left[ {{M}^{0}}L{{T}^{-2}} \right]$
The dimension of mass $\left( M \right)$ is $\left[ M{{L}^{0}}{{T}^{0}} \right]$.
The dimension of velocity $\left( V \right)$ is $\left[ {{M}^{0}}L{{T}^{-1}} \right]$.
The dimension of time $\left( T \right)$ is $\left[ {{M}^{0}}{{L}^{0}}T \right]$.
Now, let us write the dimensions of height according to the new fundamental quantities.
Hence,
$\left[ \text{height} \right]=\left[ {{A}^{w}}{{M}^{x}}{{V}^{y}}{{T}^{z}} \right]$ --(1)
where $w,x,y,z$ are some real numbers that we will find out.
Therefore, using the information that we have stated earlier, we get,
$\left[ {{M}^{0}}L{{T}^{0}} \right]={{\left[ A \right]}^{w}}{{\left[ M \right]}^{x}}{{\left[ V \right]}^{y}}{{\left[ T \right]}^{z}}$
$\therefore \left[ {{M}^{0}}L{{T}^{0}} \right]={{\left[ {{M}^{0}}L{{T}^{-2}} \right]}^{w}}{{\left[ M{{L}^{0}}{{T}^{0}} \right]}^{x}}{{\left[ {{M}^{0}}L{{T}^{-1}} \right]}^{y}}{{\left[ {{M}^{0}}{{L}^{0}}T \right]}^{z}}$
$\therefore \left[ {{M}^{0}}L{{T}^{0}} \right]=\left[ {{M}^{0}}{{L}^{w}}{{T}^{-2w}} \right]\left[ {{M}^{x}}{{L}^{0}}{{T}^{0}} \right]\left[ {{M}^{0}}{{L}^{y}}{{T}^{-y}} \right]\left[ {{M}^{0}}{{L}^{0}}{{T}^{z}} \right]$
$\therefore \left[ {{M}^{0}}L{{T}^{0}} \right]=\left[ {{M}^{0+x+0+0}}{{L}^{w+0+y+0}}{{T}^{-2w+0-y+z}} \right]$
$\therefore \left[ {{M}^{0}}L{{T}^{0}} \right]=\left[ {{M}^{x}}{{L}^{w+y}}{{T}^{-2w-y+z}} \right]$
Comparing the powers of $M$ on both sides, we get,
${{M}^{0}}={{M}^{x}}$
$\therefore x=0$ --(2)
Comparing the powers of $L$ on both sides, we get,
$L={{L}^{w+y}}$
$\therefore w+y=1$ --(3)

Comparing the powers of $T$ on both sides, we get,
${{T}^{0}}={{T}^{-2w-y+z}}$
$\therefore -2w-y+z=0$
$\therefore -w-\left( w+y \right)+z=0$
$\because -w-1+z=0$ [Using (3)]
$\therefore w=z-1$ --(4)
Using (3) and (4), we get,
$z-1+y=1$
$\therefore y=2-z$ --(5)
Now, putting (2), (4) and (5) in (1), we get,
$\left[ \text{height} \right]=\left[ {{A}^{z-1}}{{M}^{0}}{{V}^{2-z}}{{T}^{z}} \right]$
Upon analysis of the options, we see that only option D) matches this condition.
In option D), the condition is satisfied for $z=2$
Hence, the correct option is $D)\text{ }A{{T}^{2}}$

Note: Students must not think that option D) is the only correct answer to this question. It is the only correct one among the options given. However, there are infinite other correct answers, since we can take $z$ as any real number and get the dimensions in terms of the units by plugging in the powers as the functions of $z$. This will actually help the student understand why these units are not considered to be the fundamental units. The dimensional formula for a quantity should be unique which is certainly not the case when the fundamental units are changed to those given in the question.