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

Units and Measurements PYQ — JEE Advanced 2020 · Shift 1 · Q48 with Solution

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Correct answer: A, B

Let the dimensions of time and mass also be expressible in terms of $X$ : $[L]=[X^{\alpha}],\quad [LT^{-1}]=[X^{\beta}],\quad [LT^{-2}]=[X^{p}],\quad [MLT^{-1}]=[X^{q}],\quad [MLT^{-2}]=[X^{r}]$
From speed: $[LT^{-1}] = [X^{\beta}]$ Using $[L]=[X^{\alpha}]$ , we get $[T^{-1}] = [X^{\beta-\alpha}]$
From acceleration: $[LT^{-2}] = [X^{p}]$ But $[LT^{-2}] = [L][T^{-1}]^2 = [X^{\alpha}]\,[X^{\beta-\alpha}]^2 = [X^{\alpha+2\beta-2\alpha}] = [X^{2\beta-\alpha}]$ Hence, $p = 2\beta-\alpha$ or $\alpha + p = 2\beta$ So, A is correct.
Now find mass dimension in terms of $X$ .

From linear momentum: $[MLT^{-1}] = [X^{q}]$ But $[LT^{-1}] = [X^{\beta}]$ , so $[M] = [X^{q-\beta}]$
From force: $[MLT^{-2}] = [X^{r}]$ Also, $[MLT^{-2}] = [M][LT^{-2}] = [X^{q-\beta}][X^p] = [X^{p+q-\beta}]$ Hence, $r = p+q-\beta$ Rearranging, $p+q-r = \beta$ So, B is correct.
Check option C: $p-q+r = p-q+(p+q-\beta)=2p-\beta$ For this to equal $\alpha$ , we need $2p-\beta=\alpha$ Using $p=2\beta-\alpha$ , $2(2\beta-\alpha)-\beta = 3\beta-2\alpha$ This is not identically equal to $\alpha$ . So, C is incorrect.
Check option D: $p+q+r = p+q+(p+q-\beta)=2p+2q-\beta$ This is not identically equal to $\beta$ . So, D is incorrect.

Therefore, the correct options are: $\boxed{A, B}$