Question

The dimension of a/b in Relation F= a√x+bt²
where F is force, x is distance and t is time
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Answer 1

Answer:

Relation, `F=asqrt(x)+"bt"^(2)` By rules of dimensions : Dimensions of = Dimensions of `asqrt(x)` `a=(F)/(sqrt(x))=(["MLT"^(-2)])/([L^(1//2)])` `a=["ML"xx"L"^(-1//2)"T"^(-2)]` `a=["ML"^(1//2)"T"^(-2)]` dimensions of `b=(F)/(t^(2))=(["MLT"^(-2)])/(["T"^(2)])=["MLT"^(-4)]` `:.` dimensions of `a//b=(["ML"^(1//2)"T"^(-2)])/(["MLT"^(-4)])=["L"^(-1//2)"T"^(2)]`

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