Answer:
Here,
\[\text{p
}=\frac{ML{{T}^{-2}}}{{{L}^{2}}}=[M{{L}^{-1}}{{T}^{-2}}]\]
\[\text{c}=\left[
\text{L}{{\text{T}}^{-\text{1}}} \right]\]
\[\text{q }=\frac{energy}{area\times
time}=\frac{M{{L}^{2}}{{T}^{-2}}}{{{L}^{2}}T}=[M{{T}^{-3}}]\]
\[\therefore\]
\[{{\text{p}}^{\text{x}}}{{\text{q}}^{\text{y}}}{{\text{c}}^{\text{z}}}=\text{
}{{\left( \text{M}{{\text{L}}^{-\text{1}}}{{\text{T}}^{-\text{2}}}
\right)}^{\text{x}}}{{\left( \text{M}{{\text{T}}^{-\text{3}}}
\right)}^{\text{y}}}{{\left( \text{L}{{\text{T}}^{-\text{1}}}
\right)}^{\text{z}}}=\text{ }{{\text{M}}^{\text{x }+\text{
y}}}{{\text{L}}^{-\text{x }+\text{
z}}}{{\text{T}}^{-\text{2x}}}^{-\text{3y}-\text{z}}\]
It will be dimensionless, if \[\text{x}+\text{y}=0\]
\[\therefore\]
\[x=-y\]
\[\text{x
}+\text{ z }=0\]
and
\[\text{x}=\text{z}\text{2x}\text{3y}-\text{z}=0\]
These equations have infinite solutions. One of the solutions is x
= 1,y = -1 and z = 1, so that
\[{{p}^{x}}{{q}^{y}}{{c}^{z}}\]
isdimensionless.
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