Question 289: Units , Dimensions and Error Analysis

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] = [X
$$\alpha $$
]; [speed] = [X
$$\beta $$
]; [acceleration] = [X
p
]; [linear
momentum] = [X
q
]; [force] = [X
r
]. Then

Accepted Answer

$$[x] = {x^\alpha } \Rightarrow {p \over v} = time$$
$$[v] = {x^\beta }$$
$$[a] = {x^p} \Rightarrow {v \over a} = time$$
$$[p] = {x^q}$$
$$[F] = {x^r}$$
$$ \Rightarrow \left( {{{position} \over {speed}}} ight) = \left( {{{speed} \over {acceleration}}} ight) = \left( {{p \over F}} ight)$$
$$ \Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}$$
$$ \Rightarrow {x^{\alpha  - \beta }} = {x^{\beta  - p}} = {x^{q - r}}$$
$$ 	herefore $$ $$\alpha  - \beta  = \beta  - p = q - r$$
$$\alpha  + p = 2\beta $$
$$q + p = \beta  + r$$
$$p + q - r = \beta $$

Answer

$$\alpha $$ + p = 2$$\beta $$

Answer

p + q - r = $$\beta $$

Answer

p - q + r = $$\alpha $$

Answer

p + q + r = $$\beta $$

Units , Dimensions and Error Analysis

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