A) \[[{{M}^{0}}{{L}^{0}}{{T}^{-4}}]\]
B) \[[{{M}^{0}}{{L}^{0}}{{T}^{-2}}]\]
C) \[[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\]
D) \[[{{M}^{0}}{{L}^{0}}{{T}^{-3}}]\]
Correct Answer: A
Solution :
[a] : Given, \[\tau =a\times L+b\times I/\omega \] \[\therefore \]\[[a]=\frac{[\tau ]}{[L]}=\frac{[I\alpha ]}{[I\omega ]}=\frac{[{{T}^{-2}}]}{[{{T}^{-1}}]}=[{{T}^{-1}}]\] \[[b]=\frac{[\tau ]}{[I/\omega ]}=\frac{[I\alpha ]}{[I/\omega ]}=[\alpha \omega ]=[{{T}^{-2}}][{{T}^{-1}}]=[{{T}^{-3}}]\] \[\therefore \]\[[a\times b]=[{{T}^{-1}}][{{T}^{-3}}]=[{{M}^{0}}{{L}^{0}}{{T}^{-4}}]\]
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