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Check the correctness of the following equation by the method of dimensions (i) v^(2)=u^(2)+2gs (ii) s=ut+1/2"gt"^(2) (iii) v=u+"gt" |
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Answer» Solution :(i) `v^(2)=u^(2)+2gs` Dimensional formula of `v^(2)` is `[LT^(-1)]^(2)` that of `u^(2)` is `[LT^(-1)]^(2)` and gs is `[LT^(-1)]^(2)` and gs is `[LT^(-2)][L]=[L^(2)T^(-2)]` substituting in the equation `L^(2)T^(-2)=L^(2)T^(-2)+L^(2)T^(-2)` Each term on both sides of the equation has the same DIMENSIONS. So the equation is DIMENSIONALLY CORRECT. (ii) `s=ut+1/2"GT"^(2)` Dimensional formula of s is `[L]`, that of ut is `[LT^(-1)][T]=[L]` and that of `1/2"gt"^(2)` is `[LT^(-2)][T^(2)]=[L]` Substituting the dimensions in the given equations `[L]=[LT^(-1)][T]+[LT^(-2)][T^(2)]` `[L]=[L]+[L]` Eac term on both sides of the equation has the same dimensions. So the equation is dimensionally correct (III) `v=u+"gt"` Dimensional formula of v,u and gt are `[LT^(-1)],[LT^(-1)]` and `[LT^(-2)][T]` respectively. Substituting in the given equation `[LT^(-1)]=[LT^(-1)]+[LT^(-2)][T]` `=[LT^(-1)]+[LT^(-1)]` Each on both sides of equation has the same dimensions. So the equation is dimensionally correct. |
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