Explanation:
As per question
\(G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right],\)
\(h = \left[ {M{L^2}{T^{ - 1}}} \right],c = \left[ {L{T^{ - 1}}} \right]\)
Mass \(\propto h^{a} G^{b} c^{c}\)
\(\left[ {{M^1}{L^0}{T^0}} \right] = {\left[ {M{L^2}{T^{ - 1}}} \right]^a}\)
\({\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^b}{\left[ {L{T^{ - 1}}} \right]^c}\)
\(\left[ {{M^1}{L^0}{T^0}} \right] = \left[ {{M^{a - b}}{L^{2a + 3b + c}}{T^{ - a - 2b - c}}} \right]\)
\(a - b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
\(2a + 3b + c = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
\( - a - 2b - c = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
By solving, \(a=\dfrac{1}{2}, b=\dfrac{-1}{2}, c=\dfrac{1}{2}\)
\(\therefore m=\left[h^{1 / 2} c^{1 / 2} G^{-1 / 2}\right]\)
So, correct option is (3).