A) \[[{{M}^{0}}L{{T}^{-1}}]\]
B) \[[M{{L}^{0}}{{T}^{-1}}]\]
C) \[[M{{L}^{-1}}{{T}^{0}}]\]
D) \[[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
Correct Answer: C
Solution :
\[\nu =\frac{P}{2l}{{\left[ \frac{F}{m} \right]}^{1/2}}\]\[\Rightarrow {{\nu }^{2}}=\frac{{{P}^{2}}}{4{{l}^{2}}}\left[ \frac{F}{m} \right]\therefore m\propto \frac{F}{{{l}^{2}}{{\nu }^{2}}}\] \[\Rightarrow [m]=\left[ \frac{ML{{T}^{-2}}}{{{L}^{2}}{{T}^{-2}}} \right]=[M{{L}^{-1}}{{T}^{0}}]\]
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