A) \[[{{M}^{o}}{{L}^{o}}{{T}^{o}}]\]
B) \[[{{M}^{o}}{{L}^{o}}{{T}^{-1}}]\]
C) \[[{{M}^{o}}{{L}^{-1}}{{T}^{o}}]\]
D) \[[{{M}^{o}}L{{T}^{-1}}]\]
Correct Answer: D
Solution :
Given, \[F=A\,\cos \,Bx+C\,\sin \,Dt\] \[[D]=[{{T}^{-1}}]:[B]=[{{L}^{-1}}]\] \[\left[ \frac{D}{B} \right]=\left[ \frac{{{T}^{-1}}}{{{L}^{-}}} \right]=[{{M}^{0}}L{{T}^{-1}}]\]
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