A) \[ML{{T}^{-3}}\] and \[M{{L}^{2}}{{T}^{-4}}\]
B) \[ML{{T}^{-3}}\] and \[ML{{T}^{-4}}\]
C) \[ML{{T}^{-1}}\] and \[ML{{T}^{0}}\]
D) \[ML{{T}^{-4}}\] and \[ML{{T}^{1}}\]
Correct Answer: B
Solution :
From the principle of dimensional homogenity \[[a]=\left[ \frac{F}{t} \right]=[ML{{T}^{-3}}]\] and \[[b]=\left[ \frac{F}{{{t}^{2}}} \right]=[ML{{T}^{-4}}]\]
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