$\begin{array}{ll}\mathrm{We} \mathrm{know} \mathrm{that} \mathrm{dimensions} \mathrm{of} & \left(\mathrm{h}\right)=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]\\ \mathrm{Dimensions}of& \left(\mathrm{c}\right)=\left[{\mathrm{LT}}^{-1}\right]\\ \mathrm{Dimensions} \mathrm{of} \mathrm{gravitational} \mathrm{constant} \left(\mathrm{G}\right)& =\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]\end{array}$
$\left(\mathrm{i}\right) \mathbf{Step}\mathbf{ }\mathbf{2}\mathbf{:} \mathrm{Define} \mathrm{the} \mathrm{dependence} \mathrm{of} \mathrm{m} \mathrm{with} \mathrm{C}, \mathrm{h} \mathrm{and} \mathrm{G}.$
$\begin{array}{ll}\text{ Let }& \mathrm{m}\propto {\mathrm{C}}^{\mathrm{x}}{\mathrm{h}}^{\mathrm{y}}{\mathrm{G}}^{\mathrm{z}}\\ \Rightarrow & \mathrm{m}={\mathrm{kC}}^{\mathrm{x}}{\mathrm{h}}^{\mathrm{y}}\mathrm{G} ...\left(\mathrm{i}\right)\end{array}$

where k is a dimensionless constant of proportionality.

Step 3: Put the dimensions of each quantity.
Substituting dimensions of each term in Eq (i), we get,

$\begin{array}{l}\left[{\mathrm{ML}}^0{\mathrm{T}}^0\right]=[{\mathrm{LT}}^{-1}{]}^{\mathrm{x}}\times [{\mathrm{ML}}^2{\mathrm{T}}^{-1}{]}^{\mathrm{\gamma }}[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}{]}^2\\ =\left[{\mathrm{M}}^{\mathrm{y}-\mathrm{z}}{\mathrm{L}}^{\mathrm{x}+2\mathrm{y}+3\mathrm{z}}{\mathrm{T}}^{-\mathrm{x}-\mathrm{y}-2\mathrm{z}}\right]\end{array}$

Comparing powers of the same terms on both sides, we get,

y - z = 1                            ...(I)
x + 2y + 3z = 0                 ...(iii)
- x - y - 2z = 0                   ...(iii)

Adding Eqs. (i). (i) and (v), we get,
$2\mathrm{y}=1\Rightarrow \mathrm{y}=\frac{1}{2}$
Substituting the value of y in Eq. (i), we get,
$\mathrm{z}=-\frac{1}{2}$
From Eq. (iv)
x = - y - 2z
Substituting values of y and z, we get
$\mathrm{x}=-\frac{1}{2}-2(-\frac{1}{2})=\frac{1}{2}$
Putting values of x, y, and z in Eq. (i), we get

$\begin{array}{l}\mathrm{m}={\mathrm{kc}}^{1/2}{\mathrm{h}}^{1/2}{\mathrm{G}}^{-1/2}\\ \Rightarrow \mathrm{m}=\mathrm{k}\sqrt{\frac{\mathrm{ch}}{\mathrm{G}}}\end{array}$

(ii) $\mathbf{Step}\mathbf{ }\mathbf{4}\mathbf{:} \mathrm{Define} \mathrm{the} \mathrm{dependence} \mathrm{of} \mathrm{L} \mathrm{with} \mathrm{C}, \mathrm{h} \mathrm{and} \mathrm{G}.$

$\begin{array}{ll}\text{ Let }& \mathrm{L}\propto {\mathrm{c}}^{\mathrm{x}}{\mathrm{h}}^{\mathrm{y}}{\mathrm{G}}^{\mathrm{z}}\\ \Rightarrow & \mathrm{L}={\mathrm{kc}}^{\mathrm{x}}{\mathrm{h}}^{\mathrm{y}}{\mathrm{G}}^{\mathrm{z}}\end{array}$

where K is a dimensionless constant.

Step 3: Put the dimensions of each quantity.

Substituting dimensions of each term in Eq. (v), we get

$\left[{\mathrm{M}}^0{\mathrm{LT}}^0\right] = [{\mathrm{LT}}^{-1}{]}^{\mathrm{x}}\times [{\mathrm{ML}}^2{\mathrm{T}}^{-1}{]}^{\mathrm{\gamma }}\times [{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}{]}^{\mathrm{z}}$
$= \left[{\mathrm{M}}^{\mathrm{y}-\mathrm{z}}{\lfloor }^{\mathrm{x}+2\mathrm{y}+3\mathrm{z}}{\mathrm{T}}^{-\mathrm{x}-\mathrm{y}-2\mathrm{z}}\right]$

On comparing powers of same terms, we get
y - z = 0                                  ...(vi)
x + 2y + 3z = 1                       ...(vii)
- x - y - 2z = 0                         ...(viii)

(iii) Let $\mathrm{T}\propto {\mathrm{c}}^{\mathrm{a}}{\mathrm{h}}^{\mathrm{b}}{\mathrm{G}}^{\mathrm{c}}$

$\Rightarrow \mathrm{T}={\mathrm{c}}^{\mathrm{a}}{\mathrm{h}}^{\mathrm{b}}{\mathrm{G}}^{\mathrm{c}} ...........\left(\mathrm{ix}\right)$

where, k is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

$\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1\right]={\left[{\mathrm{LT}}^{-1}\right]}^{\mathrm{a}}\times {\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}^{\mathrm{b}}\times {\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]}^{\mathrm{c}}$
$=\left[{\mathrm{M}}^{\mathrm{b}-\mathrm{c}}{\mathrm{L}}^{\mathrm{a}+2\mathrm{b}+3\mathrm{c}}{\mathrm{T}}^{-\mathrm{a}-\mathrm{b}-2\mathrm{c}}\right]$

On comparing powers of the same terms, we get

$\mathrm{b}-\mathrm{c}=0 ...\left(\mathrm{x}\right)$
$\mathrm{a}+2\mathrm{b}+3\mathrm{c}=1 ...\left(\mathrm{xi}\right)$
$-\mathrm{a}-\mathrm{b}-2\mathrm{c}=1 ...\left(\mathrm{xii}\right)$

Adding Eqs. (x), (xi) and (xii), we get

$2\mathrm{b}=1\Rightarrow \mathrm{b}=\frac{1}{2}$ 

Substituting the value of b in Eq. (x), we get

$\mathrm{c}=\mathrm{b}=\frac{1}{2}$

From Eq. (xii),

$\mathrm{a}=-\mathrm{b}-2\mathrm{c}-1$

Substituting values of b and c, we get

$\mathrm{a}=-\frac{1}{2}-2\left(\frac{1}{2}\right)-1=-\frac{5}{2}$

Putting values of a, b and c in Eq. (ix), we get

$\mathrm{T}={\mathrm{kc}}^{-5/2}{\mathrm{h}}^{1/2}{\mathrm{G}}^{1/2}=\mathrm{k}\sqrt{\frac{\mathrm{hG}}{{\mathrm{c}}^5}}$