A) \[[L{{T}^{2}}],[L]\] and \[[T]\]
B) \[[{{L}^{2}}],\text{ }\!\![\!\!\text{ T }\!\!]\!\!\text{ }\] and \[[L{{T}^{2}}]\]
C) \[[L{{T}_{2}}],[LT]\] and \[[L]\]
D) \[[L],[LT]\] and \[[{{T}^{2}}]\]
Correct Answer: A
Solution :
The given expression is \[v=at+\frac{b}{t+c}\] From principle of homogeneity \[[a][t]=[v]\] \[[a]=\frac{[v]}{[t]}=\frac{[L{{T}^{-1}}]}{[T]}=[L{{T}^{-2}}]\] Similarly, \[[c]=[t]=[T]\] Further, \[\frac{[b]}{[t+c]}=[v]\] or \[[b]=[v]\,[t+c]\] or \[[b]=[L{{T}^{-1}}]\,[T]=[L]\]
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