Question

Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length 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

Hint: When a question is asked for finding the unit of a particular function or parameter, always try to find an answer using its dimension. Define each term given in question and obtain their dimensions by breaking them into fundamental physical quantities.
Students must remember the units of some basic parameters.

Complete step-by-step answer:
This question can be solved using the dimensions of Planck’s constant h, speed of light c and gravitational constant G.

We have to know dimensions of h, c and G and it is given by,

$h=\left[ M{{L}^{2}}{{T}^{-1}} \right]$ , $c=\left[ L{{T}^{-1}} \right]$ , $G=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]$
We are familiar with the and dimension of length as $L=\left[ L \right]$

Consider dimensions of length in the form of h, c and G as

$\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]={{h}^{a}}{{c}^{b}}{{G}^{d}}$ …..(1)
where, a, b and d are the indices of h, c and G respectively.

Now we put the dimensions of h, c and G in equation (1), we get,

$\begin{align}
  & \left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]={{\left[ M{{L}^{2}}{{T}^{-1}} \right]}^{a}}{{\left[ L{{T}^{-1}} \right]}^{b}}{{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]}^{d}} \\
 & \left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]=\left[ {{M}^{a-d}}{{L}^{2a+b+3d}}{{T}^{-a-b-2d}} \right] \\
\end{align}$

Comparing the indices of M, L and T, we get,

$\begin{align}
  & a-d=0 \\
 & 2a+b+3d=1 \\
 & -a-b-2d=0 \\
\end{align}$

Solving above three equations, the values of a, b and d are
$a=\dfrac{1}{2},b=\dfrac{-3}{2},d=\dfrac{1}{2}$
Hence, $L\propto {{h}^{\dfrac{1}{2}}}{{c}^{\dfrac{-3}{2}}}{{G}^{\dfrac{1}{2}}}$
Similarly, we can find the unit of mass using h, c and G.
We know the dimension of M is $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$
Here, we consider p, q and r be the indices of h, c and G respectively. So, we can write dimension of M using h, c and G as follows,

$\begin{align}
  & \left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]={{\left[ {{M}^{1}}{{L}^{2}}{{T}^{-1}} \right]}^{p}}{{\left[ L{{T}^{-1}} \right]}^{q}}{{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]}^{r}} \\
 & \left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]=\left[ {{M}^{p-r}}{{L}^{2p+q+3r}}{{T}^{-p-q-2r}} \right] \\
\end{align}$

Comparing indices of both sides, we get,

$\begin{align}
  & p-r=1 \\
 & 2p+q+3r=0 \\
 & -p-q-2r=0 \\
\end{align}$

Solving above three equations we get,

$p=\dfrac{1}{2},q=\dfrac{1}{2},r=\dfrac{-1}{2}$
Hence, $M\propto {{h}^{\dfrac{1}{2}}}{{c}^{\dfrac{1}{2}}}{{G}^{-\dfrac{1}{2}}}$
Therefore, $M\propto \sqrt{c}$ , $L\propto \sqrt{G}$ and $L\propto \sqrt{h}$
Correct options are A, C and D.

Note: While comparing the indices of both sides of equations, check it for once because there are possibilities of silly mistakes which leads to the wrong answer. So, after the solution, cross check the answer.