A) 1 D
B) -1 D
C) 25 D
D) -25 D
Correct Answer: A
Solution :
\[\text{ }\!\![\!\!\text{ }{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{1}}}\text{ }\!\!]\!\!\text{ = }\!\![\!\!\text{ }{{\text{M}}^{\text{-y+z}}}{{\text{L}}^{\text{x+3y+2z}}}{{\text{T}}^{\text{-x-2y-z}}}\text{ }\!\!]\!\!\text{ }\] \[-y+z=0\] ?.(i) and \[x+3y+2z=0\] \[-x-2y-z=1\] ?..(ii) Thus, \[x=-\frac{5}{2},y=z=\frac{1}{2}\] \[[{{G}^{1/2}}{{h}^{1/2}}{{c}^{-5/2}}]\] \[{{v}_{1}}\] \[{{v}_{2}}\] \[{{m}_{1}}{{v}_{1}}-{{m}_{2}}{{v}_{2}}=0\] \[\Rightarrow \] \[{{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}\]
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