A) \[\frac{-1}{2},\frac{1}{2}\text{and}\frac{5}{2}\]
B) \[\frac{1}{2},\frac{-1}{2}\text{and}\frac{-5}{2}\]
C) \[\frac{-1}{2},\frac{1}{2}\text{and}\frac{3}{2}\]
D) \[\frac{1}{2},\frac{-1}{2}\text{and}\frac{-3}{2}\]
Correct Answer: A
Solution :
[a]:\[E={{G}^{p}}{{h}^{q}}{{c}^{r}}\] ...(i) Equating dimensions on both sides of equation (i), we get \[[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]={{[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]}^{p}}{{[M{{L}^{2}}{{T}^{-1}}]}^{q}}{{[L{{T}^{-1}}]}^{r}}\] \[=[{{M}^{-p+q}}{{L}^{3p+2q+r}}{{T}^{-2p-q-r}}]\] Applying principle of homogeneity of dimensions, we get \[-p+q=1\] ...(ii) \[3p+2q+r=2\] ...(iii) \[-2p-q-r=-2\] ...(iv) Adding (iii) and (iv), we get \[p+q=0\] ...(v) Adding (ii) and (v), we get\[2q=1\,or\,q=\frac{1}{2}\] From (ii) \[p=q-1=\frac{1}{2}-1=-\frac{1}{2}\] Substituting the values of p and q in equation (iii), we get \[-\frac{3}{2}+1+r=2or\,r=\frac{5}{2}\] Hence,\[p=-\frac{1}{2},q=\frac{1}{2},r=\frac{5}{2}\]
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