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JEE Advanced 2020 · Units and Measurements

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]=[Xα];[] = [X^α]; [speed]=[Xβ];[] = [X^β]; [acceleration]=[Xp];[] = [X^{p}]; [linear momentum]=[Xq];[] = [X^{q}]; [force]=[Xr].] = [X^{r}]. Then (1)α+p=2β(2)p+q−r=β(3)p−q+r=α(4)p+q+r=β(1) α + p = 2β (2) p + q − r = β (3) p − q + r = α (4) p + q + r = β

Answers: A. α+p=2β\alpha + p = 2\beta · B. p+q−r=βp + q - r = \beta

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Written solution

Step 1: Define the dimensions of fundamental quantities in terms of XX.

Let the dimensions of length, time, and mass be represented in terms of the quantity XX with exponents α\alpha, τ\tau, and μ\mu respectively.

[position]=[Xα][\text{position}] = [X^{\alpha}] [time]=[Xτ][\text{time}] = [X^{\tau}] [mass]=[Xμ][\text{mass}] = [X^{\mu}]

Step 2: Express the exponents of the given physical quantities.

Using the definitions of speed, acceleration, linear momentum, and force, we write their exponents in terms of α\alpha, τ\tau, and μ\mu.

β=α−τ\beta = \alpha - \tau p=α−2τp = \alpha - 2\tau q=β+μ=α−τ+μq = \beta + \mu = \alpha - \tau + \mu r=p+μ=α−2τ+μr = p + \mu = \alpha - 2\tau + \mu

Step 3: Verify Relation 1.

Check if α+p=2β\alpha + p = 2\beta.

α+p=α+(α−2τ)\alpha + p = \alpha + (\alpha - 2\tau) α+p=2α−2τ\alpha + p = 2\alpha - 2\tau 2β=2(α−τ)=2α−2τ2\beta = 2(\alpha - \tau) = 2\alpha - 2\tau

Since both sides are equal, Relation 1 is always true.

Step 4: Verify Relation 2.

Check if p+q−r=βp + q - r = \beta.

p+q−r=p+(β+μ)−(p+μ)p + q - r = p + (\beta + \mu) - (p + \mu) p+q−r=βp + q - r = \beta

Since both sides are equal, Relation 2 is always true.

Step 5: Verify Relation 3.

Check if p−q+r=αp - q + r = \alpha.

p−q+r=(α−2τ)−(α−τ+μ)+(α−2τ+μ)p - q + r = (\alpha - 2\tau) - (\alpha - \tau + \mu) + (\alpha - 2\tau + \mu) p−q+r=α−3τp - q + r = \alpha - 3\tau

Since α−3τ≠α\alpha - 3\tau \neq \alpha in general, Relation 3 is not always true.

Step 6: Verify Relation 4.

Check if p+q+r=βp + q + r = \beta.

p+q+r=(α−2τ)+(α−τ+μ)+(α−2τ+μ)p + q + r = (\alpha - 2\tau) + (\alpha - \tau + \mu) + (\alpha - 2\tau + \mu) p+q+r=3α+2μ−5τp + q + r = 3\alpha + 2\mu - 5\tau

Since 3α+2μ−5τ≠β3\alpha + 2\mu - 5\tau \neq \beta in general, Relation 4 is not always true.

Answer: The relations that are always true are (1) α+p=2β\alpha + p = 2\beta and (2) p+q−r=βp + q - r = \beta.

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