The velocity, acceleration and force in two systems of units are related as under:
(i) `v'=(alpha^(2))/(beta)v`
(ii) `a'=(alphabeta)a`
(iii) `F'=((1)/(alphabeta))F`
All the primed symbols belong to one system and unprimed ones belong to the other system. Here `alpha and beta` are dimensionless constants. Which of the following is/are correct?

Options

A. Length standards of the two systems are related by:
`L'=((alpha^(3))/(beta^(3)))L`
B. Mass standards of the two systems are related by:
`m'=((1)/(alpha^(2)beta^(2)))m`
C. Time standards of the two systems are related by:
`T'=((alpha)/(beta^(2)))T`
D. Momentum standards of the two systems are related by:
`P'=((1)/(beta^(3)))P`

Answer

To solve the problem, we need to analyze the relationships between the quantities in the two systems of units and derive the relationships for length, mass, time, and momentum. Let's break it down step by step.

### Step 1: Analyze the given relationships

We have the following relationships:
1. \( v' = \frac{\alpha^2}{\beta} v \)
2. \( a' = \alpha \beta a \)
3. \( F' = \frac{1}{\alpha \beta} F \)

Here, \( v' \), \( a' \), and \( F' \) are quantities in one system of units, while \( v \), \( a \), and \( F \) are in another system. The constants \( \alpha \) and \( \beta \) are dimensionless.

### Step 2: Relate the length standards

From the equation for velocity:
\[ v' = \frac{L'}{T'} = \frac{\alpha^2}{\beta} \frac{L}{T} \]
This implies:
\[ \frac{L'}{T'} = \frac{\alpha^2}{\beta} \frac{L}{T} \]
Rearranging gives:
\[ L' = \frac{\alpha^2}{\beta} L T' / T \]
To find the relationship for length, we need to express \( T' \) in terms of \( T \).

### Step 3: Relate the acceleration standards

From the equation for acceleration:
\[ a' = \frac{L'}{T'^2} = \alpha \beta \frac{L}{T^2} \]
This implies:
\[ \frac{L'}{T'^2} = \alpha \beta \frac{L}{T^2} \]
Rearranging gives:
\[ L' = \alpha \beta L \frac{T'^2}{T^2} \]

### Step 4: Equate the two expressions for \( L' \)

Now we have two expressions for \( L' \):
1. \( L' = \frac{\alpha^2}{\beta} L \frac{T'}{T} \)
2. \( L' = \alpha \beta L \frac{T'^2}{T^2} \)

Setting them equal:
\[ \frac{\alpha^2}{\beta} L \frac{T'}{T} = \alpha \beta L \frac{T'^2}{T^2} \]
Cancelling \( L \) (assuming \( L \neq 0 \)):
\[ \frac{\alpha^2}{\beta} \frac{T'}{T} = \alpha \beta \frac{T'^2}{T^2} \]
Rearranging gives:
\[ \frac{T'^2}{T'} = \frac{\alpha^3}{\beta^2} T \]
Thus:
\[ T' = \frac{\alpha}{\beta^2} T \]

### Step 5: Relate the mass standards

From the equation for force:
\[ F' = \frac{M'}{T'^2} = \frac{1}{\alpha \beta} \frac{M}{T^2} \]
This implies:
\[ \frac{M'}{T'^2} = \frac{1}{\alpha \beta} \frac{M}{T^2} \]
Substituting \( T' \):
\[ M' = \frac{1}{\alpha \beta} M \left(\frac{\alpha}{\beta^2} T\right)^{-2} \]
This simplifies to:
\[ M' = \frac{1}{\alpha^2 \beta^2} M \]

### Step 6: Relate the momentum standards

Momentum \( P \) is defined as:
\[ P = MV \]
Thus:
\[ P' = M'v' = \frac{1}{\alpha^2 \beta^2} M \cdot \frac{\alpha^2}{\beta} v = \frac{1}{\beta^3} P \]

### Conclusion

From the analysis, we find:
1. Length standard: \( L' = \frac{\alpha^3}{\beta^3} L \)
2. Mass standard: \( M' = \frac{1}{\alpha^2 \beta^2} M \)
3. Time standard: \( T' = \frac{\alpha}{\beta^2} T \)
4. Momentum standard: \( P' = \frac{1}{\beta^3} P \)

### Final Answers

- Length: \( L' = \frac{\alpha^3}{\beta^3} L \) (Correct)
- Mass: \( M' = \frac{1}{\alpha^2 \beta^2} M \) (Correct)
- Time: \( T' = \frac{\alpha}{\beta^2} T \) (Correct)
- Momentum: \( P' = \frac{1}{\beta^3} P \) (Correct)