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:

$[position] = [X^{α}], [speed] = [X^{β}], [acceleration] = [X^{p}], [linear momentum] = [X^{q}], [force] = [X^{r}] .$ Then [2020]

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:

$[position] = [X^{α}], [speed] = [X^{β}], [acceleration] = [X^{p}], [linear momentum] = [X^{q}], [force] = [X^{r}] .$ Then [2020]

Smart Achievers · Accepted Answer

(1, 2)

$Given position, L = [X^{α}]$

$Speed, L T^{- 1} = [X^{β}]$

$Acceleration, L T^{- 2} = [X^{p}]$

$Linear momentum, M L T^{- 1} = [X^{q}]$

$Force, M L T^{- 2} = [X^{r}]$

$Position \div Speed = time, T = \frac{[X^{α}]}{[X^{β}]} = X^{α - β}$

$Acceleration = \frac{Speed}{Time} = \frac{X^{β}}{X^{α - β}} = X^{p}$

$X^{α - β + p} = X^{β} ∴$ $α + p = 2 β$

$Hence option (1) is correct.$

$Force = \frac{linear momentum}{time}$

$[X^{r}] = \frac{[X^{q}]}{[X^{α - β}]} \Rightarrow r = q + β - α$ $\Rightarrow r = q + β - (2 β - p)$

$\Rightarrow r = q - β + p \Rightarrow p + q - r = β$

$Hence option (2) is correct.$

Solution

1 $α + p = 2 β$

2 $p + q - r = β$

3 $p - q + r = α$

4 $p + q + r = β$

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Solution

1 α+p=2β

2 p+q-r=β

3 p-q+r=α

4 p+q+r=β

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1 $α + p = 2 β$

2 $p + q - r = β$

3 $p - q + r = α$

4 $p + q + r = β$