Some times it is convenient to construct a system of units so that quantities can be expressed in terms of a quantity X as follows : `"[position] "=[X^(alpha)],"[speed]"=[X^(beta)],"[acceleration]"=[X^(p)],"linear momentum]"=[X^(q),"[force]"=[X^(R )]`.Then

Options

A. `alpha+p=2beta`
B. `p+q-r=beta`
C. `p-q+r=alpha`
D. `p+q+r=beta`

Explanation

To solve the problem, we need to establish relationships between the quantities given in terms of the physical quantity \( X \). We will analyze the dimensions of position, speed, acceleration, linear momentum, and force in terms of their respective powers of \( X \).

### Step-by-Step Solution:

1. **Define the Relationships**:
- Position: \([position] = [X^\alpha]\)
- Speed: \([speed] = [X^\beta]\)
- Acceleration: \([acceleration] = [X^p]\)
- Linear Momentum: \([linear\ momentum] = [X^q]\)
- Force: \([force] = [X^R]\)

2. **Relate Speed to Position and Time**:
- Speed is defined as the change in position over time:
\[ [speed] = \frac{[position]}{[time]} \implies [X^\beta] = \frac{[X^\alpha]}{[T]} \]
- Rearranging gives:
\[ [X^\beta] = [X^\alpha][T^{-1}] \implies \beta = \alpha - 1 \]

3. **Relate Acceleration to Speed and Time**:
- Acceleration is defined as the change in speed over time:
\[ [acceleration] = \frac{[speed]}{[time]} \implies [X^p] = \frac{[X^\beta]}{[T]} \]
- Rearranging gives:
\[ [X^p] = [X^\beta][T^{-1}] \implies p = \beta - 1 \]

4. **Substituting for Beta**:
- From the previous step, we have \( p = (\alpha - 1) - 1 = \alpha - 2 \).

5. **Establishing the Relationship Between p, q, and R**:
- Linear momentum is defined as mass times velocity:
\[ [linear\ momentum] = [mass] \times [speed] \implies [X^q] = [M][X^\beta] \]
- Rearranging gives:
\[ [X^q] = [M][X^\beta] \implies q = \beta + 1 \]

6. **Force in Terms of Momentum and Acceleration**:
- Force is defined as the rate of change of momentum:
\[ [force] = \frac{[linear\ momentum]}{[time]} \implies [X^R] = \frac{[X^q]}{[T]} \]
- Rearranging gives:
\[ [X^R] = [X^q][T^{-1}] \implies R = q - 1 \]

7. **Final Relationships**:
- We have:
- \( \beta = \alpha - 1 \)
- \( p = \alpha - 2 \)
- \( q = \beta + 1 \)
- \( R = q - 1 \)

8. **Finding the Relationships**:
- From \( R = q - 1 \):
\[ R = (\beta + 1) - 1 = \beta \]
- Thus, we can express:
- \( R = \beta \)
- \( p = \beta - 1 \)
- \( q = \beta + 1 \)

### Conclusion:
The relationships we derived are:
- \( 2\beta = \alpha + p \)
- \( p = \beta - 1 \)
- \( q = \beta + 1 \)
- \( R = \beta \)