A) zero
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
\[y=\frac{\cos \theta {{T}^{x}}.\tau }{{{l}^{3}}}\] \[[y]=[M{{L}^{-1}}{{T}^{-2}}][T]=[{{T}^{2}}]\] \[[T]=[M{{L}^{2}}{{T}^{-2}}]\] \[\theta =\]dimensionless \[(l)=[L]\] \[[M{{L}^{-1}}{{T}^{-2}}]=\frac{{{[T]}^{x}}[M{{L}^{2}}{{T}^{-2}}]}{[{{L}^{3}}]}\] omparing the power, \[-2+x=-2\] \[x=0\]You need to login to perform this action.
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