A) \[{{\nu }^{3}}\]
B) \[{{\nu }^{4}}\]
C) \[{{\nu }^{5}}\]
D) \[{{\nu }^{6}}\]
Correct Answer: D
Solution :
[d] \[m\,\propto {{v}^{x}}{{\rho }^{y}}{{g}^{z}}\] \[M\equiv {{[L{{T}^{-1}}]}^{x}}{{[M{{L}^{-3}}]}^{y}}{{[L{{T}^{-2}}]}^{z}}\] \[{{M}^{1}}{{L}^{0}}{{T}^{0}}={{M}^{y}}{{L}^{x-3y+z}}{{T}^{-x-2z}}\] \[y=1.\,\,x-3y+z=0,-x-2z=0\] Solving above equations \[x-3-\frac{x}{2}=0\Rightarrow \frac{x}{2}=3\Rightarrow x=6\]
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