A) \[{{V}^{-2}}{{F}^{0}}{{E}^{3}}\]
B) \[{{V}^{0}}F{{E}^{2}}\]
C) \[V{{F}^{-2}}{{E}^{0}}\]
D) \[{{V}^{-2}}{{F}^{0}}E\]
Correct Answer: D
Solution :
[d] \[Let\,(M)\,={{V}^{a}}{{F}^{b}}{{E}^{c}}\] Putting the dimensions of V, F and E, we have \[(M)={{(L{{T}^{-1}})}^{a}}\times {{(ML{{T}^{-2}})}^{b}}\times {{(M{{L}^{2}}{{T}^{-2}})}^{c}}\] \[or\,{{M}^{1}}={{M}^{b+c}}{{L}^{a+b+2c}}{{T}^{-a-2b-2c}}\] Equating the powers of dimensions, we have \[b+c=1\] \[a+b+2c=0;\text{ }-a-2b-2c=0\] which give a \[=-2,\text{ }b=0\text{ }and\text{ }c=1.\] Therefore\[(M)=({{V}^{-2}}{{F}^{0}}E)\]
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