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(a)/(t^(2)) = Fv = (beta)/(x^(2)) Find ...

`(a)/(t^(2)) = Fv = (beta)/(x^(2)) ` Find dimension formula for` [a] and [beta] (here t = "time" ,F = "force", v = "velocity" , x = "distance")`

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The correct Answer is:
A, B, C, D
Since dimension of
`Fv = [Fv] = [MLT^(-2)][LT^(-1)] = (ML^(3)T^(-2)] `
so , `([beta])/([x^(2)]) = [ML^(2)T^(-3)]`
` [beta] = [ML^(4)T^(-3)] `
and `[Fv + (beta)/(x^(2)]` will also have dimension `[ML^(2)T^(-3) ] `,
so LHS should also have the same dimension ` [ML^(2)T^(-3)] `
so ` ([a])/([t^(2)]) = [ML^(2)T^(-3)] `
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