A) \[-\frac{3}{2},\,\frac{1}{2},\,1\]
B) \[-1,\,-2,\,3\]
C) \[\frac{1}{2},\,-\frac{3}{2},\,-\frac{1}{2}\]
D) \[1,\,2,\,\frac{1}{3}\]
Correct Answer: A
Solution :
By substituting the dimension of each quantity we get \[T={{[M{{L}^{-1}}{{T}^{-2}}]}^{a}}{{[{{L}^{-3}}M]}^{b}}{{[M{{T}^{-2}}]}^{c}}\] By solving we get a = ? 3/2, b = 1/2 and c = 1
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