Question

Time $\left(T\right)$, Velocity $\left(C\right)$, and angular momentum $\left(h\right)$ is chosen as fundamental quantities instead of mass, length, and time. In terms of these, the dimension of mass would be

• A. $\left[M\right]=\left[T^{-1}{C}^{-2}h\right]$
• B. $\left[M\right]=\left[T^{-1}{C}^{2}h\right]$
• C. $\left[M\right]=\left[T^{-1}{C}^{-2}{h}^{-1}\right]$
• D. $\left[M\right]=\left[T{C}^{-2}h\right]$

Answer 1

The correct option is A

$\left[M\right]=\left[T^{-1}{C}^{-2}h\right]$

Step 1: Given Data

The fundamental quantity of time $=T$

The fundamental quantity of velocity $=C$

The fundamental quantity of angular momentum $=h$

Step 2: Formula Used

The dimension of mass can be given as,

$M\propto T^x{C}^y{h}^z$

Step 3: Dimension of Mass

The dimension of mass can be given as,

$M\propto T^x{C}^y{h}^z$

We know that the dimension of mass is,

$$\begin{gathered} [M^1{L}^0{T}^0]=\left[T^x{C}^y{h}^z\right] \\ =\left[T^x\right]{\left[L{T}^{-1}\right]}^y{\left[M{L}^2{T}^{-1}\right]}^z \\ =\left[M^z{L}^{y+2z}{T}^{x-y-z}\right] \end{gathered}$$

On comparing the powers, we get,

$z=1\dots \dots \dots \dots .\left(1\right)$

$$\begin{gathered} y+2z=0 \\ \Rightarrow y+2=0 \\ \Rightarrow y=-2\dots \dots \dots .\left(2\right) \end{gathered}$$

$$\begin{gathered} x-y-z=0 \\ \Rightarrow x+2-1=0 \\ \Rightarrow x=-1\dots \dots \dots \dots .\left(3\right) \end{gathered}$$

Therefore, upon substituting we get,

$\left[M\right]=\left[T^{-1}{C}^{-2}h\right]$

Hence, the correct answer is option (A).