A) \[-\frac{5}{6},\frac{1}{2},\frac{1}{3}\]
B) \[\frac{2}{5},\frac{1}{3},\frac{1}{4}\]
C) \[\frac{4}{5},\frac{9}{8},\frac{1}{4}\]
D) \[\frac{3}{2},\frac{1}{4},\frac{5}{6}\]
Correct Answer: A
Solution :
As\[T\propto {{P}^{a}}{{\rho }^{b}}{{E}^{c}}\] From principle of homogeneity \[[T]={{[M{{L}^{-1}}{{T}^{-2}}]}^{a}}{{[M{{L}^{-3}}]}^{b}}{{[M{{L}^{2}}{{T}^{-2}}]}^{c}}\] \[[{{M}^{0}}{{L}^{0}}{{T}^{1}}]=[{{M}^{a+b+c}}{{L}^{-a-3b-2c}}{{T}^{-2a-3b-2c}}]\] Equating the exponents \[a+b+c=0,\] \[-a-3b-2c=0\] and \[-2a-3b-2c=1\] Solving these equations, we get \[a=-\frac{5}{6},b=\frac{1}{2}\]and\[c=\frac{1}{3}\]
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