Question

The energy (E), angular momentum (L), and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of the universal gravitational constant in the dimensional formula of Planck’s constant (h) is

• A. $0$
• B. $-3$
• C. $\frac{5}{3}$
• D. $1$

Answer 1

The correct option is A

$0$

Step 1: Given Data:

Energy$=E$

Angular momentum$=L$

Universal gravitational constant$=G$

Step 2: Formula Used:

Dimensional formula of energy $E=\left[M^1{L}^2{T}^{-2}\right]$

Dimensional formula of angular momentum $L=\left[M^1{L}^2{T}^{-1}\right]$

Dimensional formula of Universal gravitational constant $G=\left[M^{-1}{L}^3{T}^{-2}\right]$

Dimensional formula of Planck's constant $\left[M^1{L}^2{T}^{-1}\right]$

Step 3: Calculation of the dimension of the universal gravitational constant

Let the dimension of the universal gravitational constant is -$x$

Similarly, the dimension of angular momentum is - $y$

The dimension of energy is -$z$

Now forming the equation-

$\begin{array}{rcl}h& =& G^x{L}^y{E}^z·····\left(1\right)\end{array}$

The energy has a similar formula to work thus $F\times d$

Force has dimension $F=ma=\left[ML{T}^{-2}\right]$

Thus the dimension of energy is $E=\left[M^1{L}^2{T}^{-2}\right]$

The dimension of angular momentum can be calculated as-

$L=mvr=\left[M{L}^2{T}^{-1}\right]$

The dimension of the gravitational constant can be calculated as-

$$\begin{gathered} F=\frac{G{m}_1{m}_2}{r^2} \\ G=\frac{F{r}^2}{m_1{m}_2} \end{gathered}$$

Thus the dimension of gravitational constant obtained as-$G=\left[M^{-1}{L}^3{T}^{-2}\right]$

The dimension of Planck's constant was obtained using the formula of $E=\frac{hc}{\lambda }$

Dimensional formula of Planck's constant $\left[M^1{L}^2{T}^{-1}\right]$

Substitute the known dimensions in the equation,

$\begin{array}{rcl}\left[M^1{L}^2{T}^{-1}\right]& =& {\left[M^{-1}{L}^3{T}^{-2}\right]}^x{\left[M^1{L}^2{T}^{-1}\right]}^y{\left[M^1{L}^2{T}^{-2}\right]}^z\\ \left[M^1{L}^2{T}^{-1}\right]& =& {\left[M^1{L}^3{T}^{-2}\right]}^x{\left[M^1{L}^2{T}^{-1}\right]}^y{\left[M^1{L}^2{T}^{-2}\right]}^z\end{array}$

Compare the power of the left and right side variables,

$\begin{array}{rcl}1& =& x+y+z\dots \left(2\right)\\ 2& =& 3x+2y+2z\dots \left(3\right)\\ -1& =& -2x-y-2z\dots \left(4\right)\end{array}$

To solve the equation, Multiply the equation $\left(2\right)$ by $2$,

$2=2x+2y+2z····\left(5\right)$

Subtract equation $\left(5\right)$ from equation $\left(3\right)$,

$\begin{array}{rcl}\left(3x-2x\right)+\left(2y-2y\right)+\left(2z-2z\right)& =& 2-2\\ x& =& 0\end{array}$

So, the dimension of the universal gravitational constant would be $0$.

Hence option A is the correct answer.