JEE Advance - Physics (2020 - Paper 1 Offline - No. 12)
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] = [Xp]; [linear
momentum] = [Xq]; [force] = [Xr]. Then
$$\alpha $$ + p = 2$$\beta $$
p + q - r = $$\beta $$
p - q + r = $$\alpha $$
p + q + r = $$\beta $$
Explanation
$$[x] = {x^\alpha } \Rightarrow {p \over v} = time$$
$$[v] = {x^\beta }$$
$$[a] = {x^p} \Rightarrow {v \over a} = time$$
$$[p] = {x^q}$$
$$[F] = {x^r}$$
$$ \Rightarrow \left( {{{position} \over {speed}}} \right) = \left( {{{speed} \over {acceleration}}} \right) = \left( {{p \over F}} \right)$$
$$ \Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}$$
$$ \Rightarrow {x^{\alpha - \beta }} = {x^{\beta - p}} = {x^{q - r}}$$
$$ \therefore $$ $$\alpha - \beta = \beta - p = q - r$$
$$\alpha + p = 2\beta $$
$$q + p = \beta + r$$
$$p + q - r = \beta $$
$$[v] = {x^\beta }$$
$$[a] = {x^p} \Rightarrow {v \over a} = time$$
$$[p] = {x^q}$$
$$[F] = {x^r}$$
$$ \Rightarrow \left( {{{position} \over {speed}}} \right) = \left( {{{speed} \over {acceleration}}} \right) = \left( {{p \over F}} \right)$$
$$ \Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}$$
$$ \Rightarrow {x^{\alpha - \beta }} = {x^{\beta - p}} = {x^{q - r}}$$
$$ \therefore $$ $$\alpha - \beta = \beta - p = q - r$$
$$\alpha + p = 2\beta $$
$$q + p = \beta + r$$
$$p + q - r = \beta $$
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