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

Options

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)`, `ML^(-3)`

Answer

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 equation
The equation states that force is equal to \( \alpha \) divided by the sum of density and \( \beta^3 \). For the addition of terms to be valid, they must have the same dimensions.

### Step 2: Identify the dimensions of density
Density (\( \rho \)) is defined as mass per unit volume. Therefore, its dimensions can be expressed as:
\[ [\text{density}] = \frac{[\text{mass}]}{[\text{volume}]} = \frac{M}{L^3} \]
Thus, the dimensions of density are:
\[ [\text{density}] = M L^{-3} \]

### Step 3: Relate \( \beta^3 \) to density
Since \( \beta^3 \) must have the same dimensions as density for the addition to be valid, we can write:
\[ [\beta^3] = [\text{density}] = M L^{-3} \]
To find the dimensions of \( \beta \), we take the cube root:
\[ [\beta] = (M L^{-3})^{1/3} = M^{1/3} L^{-1} \]

### Step 4: Identify the dimensions of force
The dimensions of force (\( F \)) are given by Newton's second law, which states that force is mass times acceleration:
\[ [F] = M L T^{-2} \]

### Step 5: Relate \( \alpha \) to force and density
From the equation, we can express \( \alpha \) as:
\[ \alpha = \text{force} \times (\text{density} + \beta^3) \]
Since \( \beta^3 \) has the same dimensions as density, we can simplify this to:
\[ \alpha = \text{force} \times \text{density} \]
Thus, the dimensions of \( \alpha \) can be calculated as:
\[ [\alpha] = [F] \times [\text{density}] = (M L T^{-2}) \times (M L^{-3}) = M^2 L^{-2} T^{-2} \]

### Final Results
The dimensions of \( \alpha \) and \( \beta \) are:
- \( [\alpha] = M^2 L^{-2} T^{-2} \)
- \( [\beta] = M^{1/3} L^{-1} \)