Question

If in a hypothetical system if the angular momentum and mass are dimensionless. Then which of the following is true.

• A. The linear momentum varies as $L^{-1}$.
• B. The energy varies as $L^{-2}$.
• C. The power varies as $L^{-4}$.
• D. The force varies as $L^{-5}$.

Answer 1

Answer: (A), (B), (C)

The power varies as $L^{-4}$.

The explanation for correct option

Step 1: Given

Mass is dimensionless: $\left[M\right]=\left[M^0{L}^0{T}^0\right]$

Angular momentum is dimensionless: $\left[M{L}^2{T}^{-1}\right]=\left[M^0{L}^0{T}^0\right]$

Step 2: Formula Used

Force is given by the formula $F=ma$, where $m$ is mass and $a$ is acceleration.

Power is given by the formula $P=Fv$, where $F$ is force and $v$ is velocity.

Work done or energy is given by the formula $W=F·S$, where $F$ is force and $S$ is displacement.

Linear momentum is given by the formula $p=mv$, where $m$ is mass and $v$ is velocity.

Step 3: Solution

From $\left[M\right]=\left[M^0{L}^0{T}^0\right]$ and $\left[M{L}^2{T}^{-1}\right]=\left[M^0{L}^0{T}^0\right]$, it can be written that

$$\begin{gathered} \left[L^2{T}^{-1}\right]=\left[L^0{T}^0\right] \\ \Rightarrow \left[L^2\right]=\left[T\right] \end{gathered}$$

Option A

We know that the dimension of velocity is $\left[L{T}^{-1}\right]$. Calculate the linear momentum using the formula $p=mv$,

$\begin{array}{rcl}p& =& mv\\ & =& \left[M^0{L}^0{T}^0\right]\times \left[L{T}^{-1}\right]\\ & =& \left[M^0{L}^0{T}^0\right]\times \left[L{\left[L^2\right]}^{-1}\right]\left(\because \left[L^2\right]=\left[T\right]\right)\\ & =& \left[M^0{L}^0{T}^0\right]\times \left[L\times L^{-2}\right]\\ & =& \left[L^{-1}\right]\end{array}$

Thus, option A is correct. The linear momentum varies as $L^{-1}$.

Option B

We know that the dimension of force is $\left[ML{T}^{-2}\right]$ and displacement is $\left[L\right]$. Calculate the energy using the formula $W=F·S$,

$\begin{array}{rcl}W& =& F·S\\ & =& \left[ML{T}^{-2}\right]\times \left[L\right]\\ & =& \left[M^{0}L{\left[L^2\right]}^{-2}\right]\times \left[L\right]\left(\because \left[L^2\right]=\left[T\right]and\left[M\right]=\left[M^0\right]\right)\\ & =& \left[M^{0}L\times L^{-4}\right]\times \left[L\right]\\ & =& \left[L^{-2}\right]\end{array}$

Thus, option B is correct. The energy varies as $L^{-2}$.

Option C

We know that the dimension of force is $\left[ML{T}^{-2}\right]$ and velocity is $\left[L{T}^{-1}\right]$. Calculate the power using the formula $P=Fv$,

$\begin{array}{rcl}P& =& mv\\ & =& \left[ML{T}^{-2}\right]\times \left[L{T}^{-1}\right]\\ & =& \left[M^{0}L{\left[L^2\right]}^{-2}\right]\times \left[L{\left[L^2\right]}^{-1}\right]\left(\because \left[L^2\right]=\left[T\right]and\left[M\right]=\left[M^0\right]\right)\\ & =& \left[M^{0}L\times L^{-4}\right]\times \left[L\times L^{-2}\right]\\ & =& \left[L^{-4}\right]\end{array}$

Thus, option C is correct. The power varies as $L^{-4}$.

The explanation for incorrect option

Option D

We know that the dimension of acceleration is $\left[L{T}^{-2}\right]$. Calculate the power using the formula $F=ma$,

$\begin{array}{rcl}F& =& ma\\ & =& \left[M\right]\times \left[L{T}^{-2}\right]\\ & =& \left[M^0\right]\times \left[L{\left[L^2\right]}^{-2}\right]\left(\because \left[L^2\right]=\left[T\right]\right)\\ & =& \left[M^0\right]\times \left[L\times L^{-4}\right]\\ & =& \left[L^{-3}\right]\end{array}$

Thus, option D is incorrect. The force does not vary as $L^{-5}$.

Hence, options A, B, and C are correct.