Question

Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of L, which of the following in statement(s) is/are correct?
This question has multiple correct options.
A. The dimension of force is ${{L}^{-3}}$
B. The dimension of power is ${{L}^{-5}}$
C. The dimension of energy is ${{L}^{-2}}$
D. The dimension of linear momentum is ${{L}^{-1}}$

Answer 1

We are given a hypothetical new system of units where angular momentum and mass has no dimensions. We are given the dimension of length in the new system. In the options we are given the dimensions of force, power, energy and angular momentum in the new system. We can verify them by finding their dimensions using their equations. Thus we can find the correct answers.

Formula used:
$F=ma$
$P=Fv$
$W=F\cdot s$
$p=mv$

Complete step by step answer:
In the question it is said that in a system of units mass and angular momentum are dimensionless.
We are given the dimension of length as L.
Since mass is dimensionless in this system we can write,
$\left[ M \right]=\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]$
We know that angular momentum is given by the equation,
$l=mvr$, were ‘m’ is the mass, ‘v’ is the velocity and ‘r’ is the radius.
Therefore we will get the dimension of angular momentum as,
$\left[ l \right]=\left[ M{{L}^{2}}{{T}^{-1}} \right]$
In this new system, since angular momentum is dimensionless, we can write that
$\left[ M{{L}^{2}}{{T}^{-1}} \right]=\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]$
Since, $\left[ M \right]=\left[ {{M}^{0}} \right]$, we will get the above equation as,
$\Rightarrow \left[ {{L}^{2}}{{T}^{-1}} \right]=\left[ {{L}^{0}}{{T}^{0}} \right]$
$\Rightarrow \left[ {{L}^{2}} \right]=\left[ T \right]$
First let us calculate the dimension of force in the new system.
We now that force is given by the equation,
$F=ma$, were ‘m’ is mass and ‘a’ is acceleration.
We have the dimension of acceleration as, $\left[ a \right]=\left[ L{{T}^{-2}} \right]$
Therefore we get the dimension of force as,
$\left[ F \right]=\left[ ML{{T}^{-2}} \right]$
In this system we have
$\left[ M \right]=\left[ {{M}^{0}} \right]$
$\left[ T \right]=\left[ {{L}^{2}} \right]$
$\Rightarrow \left[ {{T}^{-2}} \right]=\left[ {{\left[ {{L}^{2}} \right]}^{-2}} \right]$
$\Rightarrow \left[ {{T}^{-2}} \right]=\left[ {{L}^{-4}} \right]$
Therefore the dimension of force will be,
$\Rightarrow \left[ F \right]=\left[ {{M}^{0}}L{{L}^{-4}} \right]$
$\Rightarrow \left[ F \right]=\left[ {{L}^{-3}} \right]$
In option A It is said that the dimension of force is, ${{L}^{-3}}$. We get the same dimension for force by calculation.
Hence option A is correct.
Now let us calculate the dimension of power.
We know that power can be written as,
$P=Fv$, where ‘F’ is force and ‘v’ is velocity.
We know that dimension of force is,
$\left[ F \right]=\left[ ML{{T}^{-2}} \right]$
And dimension of velocity is,
$\left[ v \right]=\left[ L{{T}^{-1}} \right]$
Therefore we can calculate the dimension of power as,
$\left[ P \right]=\left[ ML{{T}^{-2}} \right]\times \left[ L{{T}^{-1}} \right]$
$\Rightarrow \left[ P \right]=\left[ M{{L}^{2}}{{T}^{-3}} \right]$
From previous calculations we know that, $\left[ {{L}^{2}} \right]=\left[ T \right]$.
Therefore we can write that, $\left[ {{T}^{-3}} \right]=\left[ {{\left[ {{L}^{2}} \right]}^{-3}} \right]$
$\Rightarrow \left[ {{T}^{-3}} \right]=\left[ {{L}^{-6}} \right]$
And we also know that $\left[ M \right]=\left[ {{M}^{0}} \right]$
Therefore we will get the dimension of power as,
$\Rightarrow \left[ P \right]=\left[ {{M}^{0}}{{L}^{2}}{{L}^{-6}} \right]$
$\Rightarrow \left[ P \right]=\left[ {{L}^{-4}} \right]$
In the given option B it is said that the dimension of power is ${{L}^{-5}}$. But by calculation we get the dimension of power as ${{L}^{-4}}$.
Hence option B is incorrect.
Now let us calculate the dimension of energy or work.
We know that energy or work is the product of force and displacement. Therefore its dimension can be written as,
$\left[ W \right]=\left[ ML{{T}^{-2}} \right]\left[ L \right]$
$\Rightarrow \left[ W \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
In this system we have, $\left[ M \right]=\left[ {{M}^{0}} \right]$ and by calculation we have $\left[ T \right]=\left[ {{L}^{2}} \right]$
Therefore we get, $\Rightarrow \left[ {{T}^{-2}} \right]=\left[ {{\left[ {{L}^{2}} \right]}^{-2}} \right]$
$\Rightarrow \left[ {{T}^{-2}} \right]=\left[ {{L}^{-4}} \right]$
Thus we can calculate the energy as,
$\Rightarrow \left[ W \right]=\left[ {{M}^{0}}{{L}^{2}}{{L}^{-4}} \right]$
$\Rightarrow \left[ W \right]=\left[ {{L}^{-2}} \right]$
In option C given in the question it is said that the dimension of energy is ${{L}^{-2}}$. By calculation also we get the same value.
Hence option C is correct.
Now let us calculate the dimension for linear momentum.
We know that linear momentum is the product of mass and momentum, i.e.
$p=mv$, were ‘m’ is mass and ‘v’ is velocity.
We have the dimension of velocity as, $\left[ L{{T}^{-1}} \right]$
Since $\left[ T \right]=\left[ {{L}^{2}} \right]$, we will get,
$\Rightarrow \left[ {{T}^{-1}} \right]=\left[ {{\left[ {{L}^{2}} \right]}^{-1}} \right]$
$\Rightarrow \left[ {{T}^{-1}} \right]=\left[ {{L}^{-2}} \right]$
And we have $\left[ M \right]=\left[ {{M}^{0}} \right]$
Therefore the dimension of linear momentum will be,
$\Rightarrow \left[ p \right]=\left[ {{M}^{0}}L{{L}^{-2}} \right]$
$\Rightarrow \left[ p \right]=\left[ {{L}^{-1}} \right]$
Thus we get the dimension of linear momentum as, ${{L}^{-1}}$. In option D also it is said that the dimension of linear momentum is ${{L}^{-1}}$.
Therefore option D is also correct.
Hence the correct answers are option A, C and D.

Note:
Dimension of a physical quantity is simply defined as the powers to which the fundamental units of the unit system are raised to get one unit of that quantity. Principle of homogeneity is a principle that allows us to analyze the dimensions. The principle states that the dimension of each term on the left hand side of the dimensional equation will be the same as the dimension of every term on the right hand side of that equation.