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( alpha)/( t^(2)) = Fv + ( beta)/(x^(2))...

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

Text Solution

Verified by Experts

The correct Answer is:
`[beta] = M^(1) L^(4) T^(-3) ; [alpha] = M^(1) L^(2) T^(-1)`
Since `[Fv] = M^(1) L^(2) T^(-3)`,
So `[ (beta)/( x^(2))]` should also be `M^(1) L^(2)T^(-3)`
`[beta] = M^(1) L^(4) T^(-3) and [Fv + ( beta)/( x^(2))]` will also have dimension
`M^(1)L^(2) T^(-3)`. So,
`([alpha)]/([ t^(2)]) = M^(1) L^(2)T^(-3) , [alpha] = M^(1) L^(2) T^(-1)`
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Knowledge Check

  • Given: Force = (alpha)/("density " + beta^2) .What are the dimensions of alpha , beta ?

    A
    `ML^(-2) T^(-2) , ML^(1//2)`
    B
    `M^2 L^4 T^(-2), M^(-1//3) L^(-1)`
    C
    `M^2 L^(-2) T^(-2) , M^(1//3) L^(-1)`
    D
    `M^2 L^(-2) T^(-2), ML^(-3)`

  • Given : force = (alpha)/("density" + beta^3). What are the dimensions of alpha, beta ?

    A
    `M L^(-2) T^(-2), ML^(-1//3)`
    B
    `M^2 L^4 T^(-2), M^(1//3) L^(-1)`
    C
    `M^2 L^(-2) T^(-2), M^(1//3) L^(-1)`
    D
    `M^2 L^(-2), M L^(-3)`

  • Given force =(alpha)/("density"+beta^(3)) what are dimensions of alpha, beta ?

    A
    `ML^(–2) T^(–2) , ML^(–1//3)`
    B
    `M^(2) L^(4) T^(–2), M^(1//3) L^(–1)`
    C
    `M^(2) L^(–2) T^(–2), M^(1//3) L^(–1)`
    D
    `M^(2) L^(–2) T^(–2), M L^(–3)`

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