Verified Solution • 2025–26 Session moderate

QJEEAdv20Phfa27: JEE Advanced 2020 Physics - Class 11 JEEAdv20Phfa27 Solution: Sometimes it is convenient to construct a system of units so…

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\alpha ]; [speed] = [X β\beta ]; [acceleration] = [Xp]; [linear momentum] = [Xq]; [force] = [Xr]. Then

Multiple Choice Options

  • A α\alpha + p = 2 β\beta
  • B p + q - r = β\beta
  • C p - q + r = α\alpha
  • D p + q + r = β\beta

Step-by-Step Verified Solution

[x]=xα⇒pv=time[x] = {x^\alpha } \Rightarrow {p \over v} = time

[v]=xβ[v] = {x^\beta }

[a]=xp⇒va=time[a] = {x^p} \Rightarrow {v \over a} = time

[p]=xq[p] = {x^q}

[F]=xr[F] = {x^r}

⇒(positionspeed)=(speedacceleration)=(pF)\Rightarrow \left( {{{position} \over {speed}}} \right) = \left( {{{speed} \over {acceleration}}} \right) = \left( {{p \over F}} \right)

⇒xαxβ=xβxα=xqxr\Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}

⇒xα−β=xβ−p=xq−r\Rightarrow {x^{\alpha - \beta }} = {x^{\beta - p}} = {x^{q - r}}

∴\therefore α−β=β−p=q−r\alpha - \beta = \beta - p = q - r

α+p=2β\alpha + p = 2\beta

q+p=β+rq + p = \beta + r

p+q−r=βp + q - r = \beta

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