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 $X$ as follows: [position] = $\left[X^{\alpha }\right]$; [speed] = $\left[X^{\beta }\right]$; [acceleration] =$\left[X^p\right]$; [linear momentum] = $\left[X^q\right]$; [force] = $\left[X^r\right]$Then

• A. $\alpha +p=2\beta$
• B. $p+q-r=\beta$
• C. $p-q+r=\alpha$
• D. $p+q+r=\beta$

Answer 1

Answer: (A), (B)

$p+q-r=\beta$

Step 1: Given data:

Position= $\left[X^{\alpha }\right]$

Speed= $\left[X^{\beta }\right]$

Acceleration=$\left[X^p\right]$

Linear momentum = $\left[X^q\right]$

Force = $\left[X^r\right]$

Step 2: Formula used:

The dimensional formula of position is $\left[L\right]$

The dimensional formula of speed is $\left[L{T}^{-1}\right]$

The dimensional formula of acceleration is $\left[L{T}^{-2}\right]$

The dimensional formula of linear momentum is $\left[ML{T}^{-1}\right]$

The dimensional formula of force is $\left[ML{T}^{-2}\right]$

Step 3: Equating the equation:

Equating position dimensions-

$\left[L\right]$ = $\left[X^{\alpha }\right]$…………(1)

Equating speed dimensions-

$\left[L{T}^{-1}\right]=$$\left[X^{\beta }\right]$…………(2)

Equating acceleration dimensions-

$\left[L{T}^{-2}\right]=$$\left[X^p\right]$…………(3)

Equating linear momentum dimensions-

$\left[ML{T}^{-1}\right]=$$\left[X^q\right]$…………(4)

Equating force dimensions-

$\left[ML{T}^{-2}\right]=$$\left[X^r\right]$…………(5)

Step 4: Solving the equations:

First divide equation (1) by equation (2)-

$\left[T\right]=\left[X^{\alpha -\beta }\right]$…………(6)

From equation (3) and (6)

$\left[L{T}^{-2}\right]=$$\left[X^p\right]$

$\frac{X^{\alpha }}{X^{2\left(\alpha -\beta \right)}}=X^p$

$$\begin{gathered} X^{\alpha -2\alpha +2\beta }=X^p \\ Onequating, \\ -\alpha +2\beta =p \\ \alpha +p=2\beta \end{gathered}$$

Thus, option A is the correct answer.

From equation (4)-

$M=X^{q-\beta }$

From equation (5)-

$$\begin{gathered} X^{q-\beta }{X}^p=X^r \\ q-\beta +p=r \\ p+q-r=\beta \end{gathered}$$

Thus, option B is also correct.

Hence, both options A, B are correct.