Answer:
Let \[m=k{{c}^{x}}{{h}^{y}}{{G}^{z}}\] \[\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]={{\left[ L{{T}^{-1}} \right]}^{x}}{{\left[ M{{L}^{2}}{{T}^{-1}} \right]}^{y}}{{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]}^{z}}\] ... (i) \[={{M}^{y-z}}.{{L}^{x+2y+3z}}{{T}^{-x-y-2z}}\] Applying the principle of homogeneity of dimensions. \[y-z=1\] ... (ii) \[x+2y+3z=0\] ... (iii) \[-x-y-2z=0\] ... (iv) Adding (iii) and (iv) we get \[y+z=0\] Adding (ii) and (v) \[2y=1,\,y=1/2\] From (v), \[\text{z }=-y=-\text{1}/\text{2}\] From (iii), \[x=-2y-3z=-1+3/2=1/2\] \[\therefore \]from (i), \[m=k{{c}^{1/2}}{{h}^{1/2}}{{G}^{-1/2}}=k\sqrt{\frac{ch}{G}}\] Taking \[k=1\]; \[m=\sqrt{\frac{3\times {{10}^{8}}\times 6.6\times {{10}^{-34}}}{6.67\times {{10}^{-11}}}}\approx {{10}^{-7}}kg\]
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