Answer:
Let
\[\text{T}=\text{K}{{\text{R}}^{\text{a}}}{{\rho
}^{\text{b}}}{{\text{G}}^{\text{c}}}\]
..(i)
\[\left[
{{\text{M}}^{0}}{{\text{L}}^{0}}{{\text{T}}^{\text{1}}} \right]={{\left[
\text{L} \right]}^{\text{a}}}{{\left[ \text{M}{{\text{L}}^{-\text{3}}}
\right]}^{\text{b}}}{{\left[
{{\text{M}}^{-\text{1}}}{{\text{L}}^{\text{3}}}{{\text{T}}^{-\text{2}}}
\right]}^{\text{c}}}={{\text{M}}^{\text{b}-\text{c}}}{{\text{L}}^{\text{a}-\text{3b}+\text{3c}}}{{\text{T}}^{-\text{2c}}}\]
Applying principle of homogeneity ef dimensions, we get
\[\text{b}\text{c}=0,\text{
a}\text{3b}+\text{3c}=0,\text{2c}=\text{l},\text{ c}=\frac{1}{2}\]
\[\text{b}=\text{c
}=\frac{1}{2}\]
and
\[\text{a}\text{3}\left(
-\frac{1}{2} \right)+\text{3}\left( -\frac{1}{2} \right)=0,\text{ a}=0\]
Putting in (i), we get
\[\text{T}=\text{KR}{}^\circ {{\rho
}^{-\text{1}/\text{2}}}{{\text{G}}^{-\text{1}/\text{2}}}=\text{K}{{(\rho
\text{G})}^{-\text{1}/\text{2}}}\]
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