Question

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity willows [position] [X"] [speed] [X] [acceleration] [X] [linear momentum] [X] [force] [X] Then which one is correct from the following two-

• A. γδέττα
• B. γδ-r-β

Answer 1

Both α + p = 2β and p + q - r = β are correct.

Answer 2

The problem asks to identify correct relationships between exponents (α, β, p, q, r) representing the dimensions of physical quantities in a system where all quantities are expressed in terms of a single quantity 'X'.

Let the dimensions of Length, Mass, and Time be [L], [M], and [T] respectively. The given dimensional relations are:

1. [Position] = [L] = [X^α]
2. [Speed] = [L T⁻¹] = [X^β]
3. [Acceleration] = [L T⁻²] = [X^p]
4. [Linear momentum] = [M L T⁻¹] = [X^q]
5. [Force] = [M L T⁻²] = [X^r]

We can derive relationships between the exponents using the fundamental definitions of these physical quantities.

Derivation 1: Relationship between α, β, and p

• We know that Speed is Position divided by Time: [Speed] = [Position] / [Time] [X^β] = [X^α] / [T] Therefore, [T] = [X^α] / [X^β] = [X^(α-β)] (Equation A)

• We also know that Acceleration is Position divided by Time squared: [Acceleration] = [Position] / [Time]² [X^p] = [X^α] / [T]² Therefore, [T]² = [X^α] / [X^p] = [X^(α-p)] Taking the square root, [T] = [X^((α-p)/2)] (Equation B)

• Equating the expressions for [T] from (A) and (B): [X^(α-β)] = [X^((α-p)/2)] Comparing the exponents: α - β = (α - p) / 2 2(α - β) = α - p 2α - 2β = α - p Rearranging the terms, we get: α + p = 2β

Derivation 2: Relationship between p, q, r, and β

• We know that Force is Mass times Acceleration: [Force] = [Mass] × [Acceleration] [X^r] = [Mass] × [X^p] Therefore, [Mass] = [X^r] / [X^p] = [X^(r-p)] (Equation C)

• We also know that Linear momentum is Mass times Speed: [Linear momentum] = [Mass] × [Speed] [X^q] = [Mass] × [X^β] Therefore, [Mass] = [X^q] / [X^β] = [X^(q-β)] (Equation D)

• Equating the expressions for [Mass] from (C) and (D): [X^(r-p)] = [X^(q-β)] Comparing the exponents: r - p = q - β Rearranging the terms, we get: p + q - r = β

The question provides options as "(1) γδέττα" and "(11) γδ-r-β", which are symbolic representations. Based on dimensional analysis, the two correct relationships that can be derived are:

1. α + p = 2β
2. p + q - r = β

Both of these relationships are dimensionally consistent and correct.

The final answer is $\boxed{\text{Both \alpha + p = 2\beta \text{ and } p + q - r = \beta \text{ are correct.}}}$

Explanation of the solution: By expressing the fundamental dimensions (Length, Mass, Time) in terms of the given quantity 'X' and its exponents (α, β, p, q, r), we use the definitions of physical quantities (speed = position/time, acceleration = position/time², force = mass × acceleration, momentum = mass × speed) to establish relationships between these exponents. This leads to two correct relationships: α + p = 2β and p + q - r = β.

Answer: The correct relationships are:

1. $\alpha + p = 2\beta$
2. $p + q - r = \beta$