Question

Planck's constant $h$, speed of light $c$ and gravitational constant $G$ are used to form a unit of length $L$ and a unit of mass $M$. Then the correct option(s) is/are

• A. $M\propto \sqrt{c}$
• B. $M\propto \sqrt{G}$
• C. $L\propto \sqrt{h}$
• D. $L\propto \sqrt{G}$

Answer 1

Answer: (A), (C), (D)

$L\propto \sqrt{G}$

Let
$L=h^a{c}^b{G}^d$ .....(1)
As we know,
$h=M{L}^2{T}^{-1}$
$c=L{T}^{-1}$
$G=M^{-1}{L}^3{T}^{-2}$

for lenght calculation,

$\left[M^0{L}^1{T}^0\right]=\left[M{L}^2{T}^{-1}{]}^a\right[L{T}^{-1}{]}^b[M^{-1}{L}^3{T}^{-2}{]}^d$
$\left[M^0{L}^1{T}^0\right]=M^a{L}^{2a}{T}^{-a}{L}^b{T}^{-b}{M}^{-d}{L}^{3d}{T}^{-2d}$
Equating coefficients of $M$, $L$ and $T$, by comparing LHS to RHS
$a-d=0,2a+b+3d=1,-a-b-3d=0$
$\Rightarrow$ $a=\frac{1}{2},b=\frac{-3}{2},b=\frac{1}{2}$
On putting $a,b$ and $d$ value at eq.(1)
$L=h^{1/2}{C}^{-3/2}{G}^{1/2}$
using above expression,
$L\propto \sqrt{h}$ and $L\propto \sqrt{G}$
so option $C$ and $D$ are correct.

Similarly using dimenitional analysis for mass,
$M=h^a{c}^b{G}^d$
$\left[M^1{L}^0{T}^0\right]=\left[M{L}^2{T}^{-1}{]}^a\right[L{T}^{-1}{]}^b[M^{-1}{L}^3{T}^{-2}{]}^d$
$M^1{L}^0{T}^0={M^{a}L}^{2a}{T}^{-a}{L}^b{T}^{-b}{M}^{-d}{L}^{3d}{T}^{-2d}$
$a-d=1,2a+b+3d=0,-a-b-2d=0$
on solving further,
$a=\frac{1}{2},b=\frac{1}{2},d=\frac{-1}{2}M\propto \sqrt{h},M\propto 1/\sqrt{G},M\propto \sqrt{c}$

Correct options are A, C and D