A) \[\left[ {{M}^{0}}L{{T}^{-1}} \right]\]and\[\left[ L{{T}^{-1}} \right]\]
B) \[\left[ {{M}^{0}}L{{T}^{-1}} \right]and\left[ {{T}^{-1}} \right]\]
C) \[\left[ {{M}^{0}}L{{T}^{-1}} \right]and\,\left[ T \right]\]
D) \[\left[ {{M}^{0}}L{{T}^{0}} \right]and\,\left[ {{T}^{-1}} \right]\]
Correct Answer: B
Solution :
Dimension of\[\left[ \frac{{{v}_{0}}}{\alpha } \right]=\] Dimension of \[x\] and dimension of \[\alpha =\frac{1}{Dimension\,of\,t}\] \[=\frac{1}{[T]}\] \[=[{{T}^{-1}}]\] Dimensions of\[{{v}_{0}}=[{{M}^{0}}L{{T}^{-1}}]\]You need to login to perform this action.
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