A)
a b c \[[{{L}^{2}}]\] \[[T]\] \[[L{{T}^{2}}]\]
B)
a b c \[[L{{T}^{2}}]\] \[[LT]\] \[[T]\]
C)
a b c \[[L{{T}^{-2}}]\] \[[L]\] \[[T]\]
D)
a b c \[[L]\] \[[LT]\] \[[{{T}^{2}}]\]
Correct Answer: C
Solution :
Given, \[v=at+\frac{b}{t+c}\] or \[[at]\,=[v]=[L{{T}^{-1}}]\] \[\therefore \] \[[a]=\frac{[L{{T}^{-1}}]}{[T]}=[L{{T}^{-2}}]\] Dimension \[c=[t]=[T]\] (we can add quantities of same dimensions only). \[\left[ \frac{b}{t+c} \right]=[v]\,=[L{{T}^{-1}}]\] or \[[b]=[L{{T}^{-1}}]\,[T]=[L]\]
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