Answer:
Let
\[l=k{{G}^{x}}{{c}^{y}}{{h}^{z}}\]
\[\left[
{{\text{M}}^{0}}{{\text{L}}^{\text{1}}}{{\text{T}}^{0}} \right]\text{ }=\text{
}{{\left[
{{\text{M}}^{-\text{1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{-\text{2}}}
\right]}^{\text{x}}}\left[ \text{L}{{\text{T}}^{-\text{1}}} \right]{{\left[
\text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{-\text{1}}} \right]}^{\text{2}}}\]
=
\[=[{{\text{M}}^{-}}^{\text{x}+\text{z}}{{\text{L}}^{\text{3x}+\text{y
2 z}}}{{\text{T}}^{\text{y}+}}^{\text{2z}}{{\text{T}}^{-\text{2x }-\text{ y
}-\text{ z}}}]\]
...(i)
Applying the principle of homogeneity of dimensions, we get
\[-\text{x}+\text{z}=0\]
,
\[\text{3x}+\text{y}+\text{2z=1}\]
,
\[-2x-y-z=0\]
.
Solving these eqns., we get
\[x=\frac{1}{2}\]
,
\[\,y=\frac{-3}{2}\]
,
\[\,z=\frac{1}{2}\]
\[\therefore\]
From(i),
\[l=\text{k}{{\text{G}}^{\text{1}/\text{2}}}{{\text{c}}^{-\text{3}/\text{2}}}{{\text{h}}^{\text{1}/\text{2}}}=\text{
k}\sqrt{\frac{Gh}{{{c}^{3}}}}\]
Taking
\[\text{k}=\text{1};\text{
}\]
\[\text{l }=\sqrt{\frac{(6.67\times
{{10}^{-11}})(6.63\times {{10}^{-34}})}{{{(3\times {{10}^{8}})}^{3}}}}\approx
\text{1}{{0}^{-\text{35}}}\text{m}\]
.
This length is called Planck's length.
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