A) \[F{{A}^{-1}}v\]
B) \[F{{v}^{3}}{{A}^{-2}}\]
C) \[F{{v}^{2}}{{A}^{-1}}\]
D) \[{{F}^{2}}{{v}^{2}}{{A}^{-1}}\]
Correct Answer: B
Solution :
\[L\propto {{v}^{x}}{{A}^{y}}{{F}^{z}}\] \[\Rightarrow \] \[L=k{{v}^{x}}{{A}^{y}}{{F}^{z}}\] Putting the dimensions in the above relation \[[M{{L}^{2}}{{T}^{-1}}]=k{{[L{{T}^{-1}}]}^{x}}{{[L{{T}^{-2}}]}^{y}}{{[ML{{T}^{-2}}]}^{z}}\] Þ \[[M{{L}^{2}}{{T}^{-1}}]=k[{{M}^{z}}{{L}^{x+y+z}}{{T}^{-x-2y-2z}}]\] Comparing the powers of \[M,\,L\] and \[T\] \[z=1\] ?(i) \[x+y+z=2\] ?(ii) \[-x-2y-2z=-1\] ?(iii) On solving (i), (ii) and (iii) \[x=3,\,y=-2,\,z=1\] So dimension of \[L\] in terms of \[v,\,A\] and \[f\] \[[L]=[F{{v}^{3}}{{A}^{-2}}]\]
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