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Given : force = (alpha)/("density" + bet...

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)`

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AI Generated Solution

The correct Answer is:
To find the dimensions of \( \alpha \) and \( \beta \) in the equation \( \text{force} = \frac{\alpha}{\text{density} + \beta^3} \), we will follow these steps: ### Step 1: Understand the dimensions of force and density - The dimension of force \( F \) is given by: \[ [F] = [M][L][T]^{-2} = MLT^{-2} \] - The dimension of density \( \rho \) is given by: \[ [\rho] = \frac{[M]}{[L]^3} = ML^{-3} \] ### Step 2: Analyze the equation The equation can be rewritten as: \[ F = \frac{\alpha}{\rho + \beta^3} \] For the equation to be dimensionally consistent, the dimensions of \( \alpha \) must match the dimensions of \( F \) when multiplied by the dimensions of \( \rho + \beta^3 \). ### Step 3: Determine the dimensions of \( \beta \) Since \( \beta^3 \) must have the same dimensions as density \( \rho \): \[ [\beta^3] = [\rho] = ML^{-3} \] Thus, we can write: \[ [\beta]^3 = ML^{-3} \] Taking the cube root gives us: \[ [\beta] = M^{1/3}L^{-1} \] ### Step 4: Substitute \( \beta \) back into the equation Now we can substitute \( \beta \) back into the equation to find \( \alpha \). Since \( \rho + \beta^3 \) has the same dimensions as \( \rho \) (as \( \beta^3 \) is of the same dimension): \[ [\rho + \beta^3] = ML^{-3} \] ### Step 5: Rearranging the equation for \( \alpha \) From the equation \( F = \frac{\alpha}{\rho + \beta^3} \), we can rearrange it to find \( \alpha \): \[ \alpha = F \cdot (\rho + \beta^3) \] Substituting the dimensions we have: \[ [\alpha] = [F] \cdot [\rho] = (MLT^{-2}) \cdot (ML^{-3}) = M^2L^{-2}T^{-2} \] ### Final Dimensions - The dimensions of \( \alpha \) are \( M^2L^{-2}T^{-2} \). - The dimensions of \( \beta \) are \( M^{1/3}L^{-1} \). ### Summary - Dimensions of \( \alpha \): \( M^2L^{-2}T^{-2} \) - Dimensions of \( \beta \): \( M^{1/3}L^{-1} \)

To find the dimensions of \( \alpha \) and \( \beta \) in the equation \( \text{force} = \frac{\alpha}{\text{density} + \beta^3} \), we will follow these steps: ### Step 1: Understand the dimensions of force and density - The dimension of force \( F \) is given by: \[ [F] = [M][L][T]^{-2} = MLT^{-2} \] - The dimension of density \( \rho \) is given by: ...
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