If velocity of light $c$, Planck's constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.

Options:

None

Solution:
As we know that, principle of homogeneity of dimensions that is LHS and RHS of an equation will have same dimensions.
We also know that, dimensions of $(h)$
$$
=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
$$
$\therefore$ Dimensions of $(c)=\left[\mathrm{LT}^{-1}\right]$
Dimensions of gravitation constant $(G)$
$$
\begin{aligned}
&=\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]\left[\mathrm{M}^{-2}\right] \\
&=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]
\end{aligned}
$$
(i) Let we consider mass, $M \propto c^x h^y G^z$
$$
\Rightarrow M=k c^x h^y G^z
$$
where, $k$ is a dimensionless constant of proportianlity. Substituting dimensions of each term in Eq. (i), we get
$$
\begin{aligned}
{\left[\mathrm{ML}^0 \mathrm{~T}^0\right] } &=\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\
&=\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right]
\end{aligned}
$$
By principle of homogeneity.
Comparing powers of same terms on both sides, we have
$$
\begin{aligned}
&y-z=1 \\
&x+2 y+3 z=0 \\
&-x-y-2 z=0
\end{aligned}
$$
Adding Eqs. (ii), (iii) and (iv), we have
$$
2 y=1 \Rightarrow y=\frac{1}{2}
$$
Substituting value of $y$ in Eq. (ii), we have
$$
z=-\frac{1}{2}
$$
From Eq. (iv)
$$
x=-y-2 z
$$
Substituting values of $y$ and $z$, we have
$$
x=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2}
$$
Putting values of $x, y$ and $z$ in Eq. (i), we have
$$
\begin{aligned}
&m=k c^{1 / 2} h^{1 / 2} G^{-1 / 2} \\
&\Rightarrow m=k \sqrt{\frac{c h}{G}}
\end{aligned}
$$
(ii) Let us consider $(L)$,
$$
\begin{gathered}
L \propto c^x h^y G^z \\
\Rightarrow L=k c^x h^y G^z \ldots \text { (i) }
\end{gathered}
$$
where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. (v), we get
$$
\begin{aligned}
{\left[\mathrm{M}^0 \mathrm{LT}^0\right] } &=\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right\}^y \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\
&=\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{x-y-2 z}\right]
\end{aligned}
$$
By the principle of homogeneity comparing powers of same terms M, L and T, we get
$$
\begin{array}{ll}
y-z=0 & \text {...(ii) } \\
x+2 y+3 z=1 & \text {...(iii) } \\
-x-y-2 z=0 & \text {...(iv) }
\end{array}
$$
Adding Eqs. (ii), (iii) and (iv), we have
$$
2 y=1
$$
So, $y=\frac{1}{2}$
Substituting value of $y$ in Eq. (iii), we have
$$
z=\frac{1}{2}
$$
From Eq. (iv),
$$
x=-y-2 z
$$
Substituting values of $y$ and $z$, we have
$$
x=-\frac{1}{2}-2\left(\frac{1}{2}\right)=-\frac{3}{2}
$$
Putting value of $x, y$ and $z$ in Eq. (i), we have
$$
\begin{aligned}
&L=k c^{-3 / 2} h^{1 / 2} G^{1 / 2} \\
&L=k \sqrt{\frac{h G}{c^3}}
\end{aligned}
$$
(iii) Let us consider, $T \propto c^x h^y G^z$
So, $T=k c^x h^y G^z$...(i)
where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. (i), we get
$$
\begin{aligned}
{\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right] } &=\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y \\
& \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right] z \\
&=\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right]
\end{aligned}
$$
By the principle of homogeneity comparing powers of same terms, as $M, L$ and $T$ so we get
$$
\begin{aligned}
&y-z=0 \\
&x+2 y+3 z=0 \\
&-x-y-2 z=1
\end{aligned}
$$
Adding Eqs. (ii), (iii) and (iv) we have
$$
2 y=1
$$
So, $y=\frac{1}{2}$
Substituting value of $y$ in Eq. (ii), we have
$$
z=y=\frac{1}{2}
$$
From Eq. (iv)
$$
x=-y-2 z-1
$$
Substituting values of $y$ and $z$, we have
$$
x=-\frac{1}{2}-2\left(\frac{1}{2}\right)-1=-\frac{5}{2}
$$
Putting values of $x, y$ and $z$ in Eq. (i), we get
$$
\begin{aligned}
&T=k c^{-5 / 2} h^{1 / 2} G^{1 / 2} \\
&T=k \sqrt{\frac{h G}{c^5}}
\end{aligned}
$$

Topic: Physics, Units and Dimensions