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Answer (3)
Given, x=at+bt2−ct3x=at+bt^2-ct^3x=at+bt2−ct3 ∴[ct3]=[x]\displaystyle \therefore [ct^3]=[x]∴[ct3]=[x] ⇒[c]=[x][t3]=[L][T3]=[LT−3]\displaystyle \Rightarrow [c]=\frac {[x]} {[t^3]} =\frac {[L]} {[T^3]}=[LT^{-3}] ⇒[c]=[t3][x]=[T3][L]=[LT−3]
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