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