A) 1
B) 0
C) 2
D) 3
Correct Answer: B
Solution :
We know that \[[Y]=[M{{L}^{-1}}{{T}^{-2}}],\,\,\sin \theta =[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[[I]=[L]\] Now writing dimensions, we have \[[M{{L}^{-1}}{{T}^{-2}}]=\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}][T]\phi [M{{L}^{2}}{{T}^{-2}}]}{[{{L}^{4}}]}\]comparing LHS and RHS, we have \[\phi =0\]You need to login to perform this action.
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