A) \[\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{\text{-1}}}\text{ }\!\!]\!\!\text{ }\,\text{and}\,\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{\text{0}}}\text{ }\!\!]\!\!\text{ }\]
B) \[\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{\text{-3}}}\text{ }\!\!]\!\!\text{ }\,\text{and}\,\text{ }\!\![\!\!\text{ M}{{\text{L}}^{2}}{{\text{T}}^{4}}\text{ }\!\!]\!\!\text{ }\]
C) \[\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{\text{-4}}}\text{ }\!\!]\!\!\text{ }\,\text{and}\,\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{1}}\text{ }\!\!]\!\!\text{ }\]
D) \[\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{\text{-3}}}\text{ }\!\!]\!\!\text{ }\,\text{and}\,\text{ }\!\![\!\!\text{ ML}{{\text{T}}^{-4}}\text{ }\!\!]\!\!\text{ }\]
Correct Answer: D
Solution :
Force \[F=at+b{{t}^{2}}\] From principle of homogeneity, Dimensions of \[at\]= dimensions of \[F\] \[\therefore \] Dimensions of \[a=\frac{[F]}{[t]}=\frac{[ML{{T}^{-2}}]}{[T]}\] \[=[ML{{T}^{-3}}]\] Similarly, dimensions of \[b=\frac{[F]}{[{{t}^{2}}]}=\frac{[ML{{T}^{-2}}]}{[{{T}^{2}}]}=[ML{{T}^{-4}}]\]
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