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(A) Let the dimensions of the fundamental quantity $X$ be $[X] = [M^a L^b T^c]$.Given:$[L] = [X^\alpha] = [M^{a\alpha} L^{b\alpha} T^{c\alpha}] \impli

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$A, B$

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(A) Let the dimensions of the fundamental quantity $X$ be $[X] = [M^a L^b T^c]$.Given:$[L] = [X^\alpha] = [M^{a\alpha} L^{b\alpha} T^{c\alpha}] \implies a\alpha = 0, b\alpha = 1, c\alpha = 0 \implies a=0, c=0, b=1/\alpha$.$[LT^{-1}] = [X^\beta] = [M^{a\beta} L^{b\beta} T^{c\beta}] \implies b\beta = 1, c\beta = -1 \implies \beta = \beta(b) = 1/\alpha \implies \beta = 1/\alpha \implies b\beta = 1$.From $[LT^{-2}] = [X^p]$, we have $b p = 1$ and $c p = -2$. Since $b = 1/\alpha$, $p = \alpha$. Also $c = -2/p = -2/\alpha$.From $[MLT^{-1}] = [X^q]$, we have $a q = 1, b q = 1, c q = -1$.From $[MLT^{-2}] = [X^r]$, we have $a r = 1, b r = 1, c r = -2$.Using these relations, we find:$1$) $\alpha + p = 2\beta$ is correct because $[L] [LT^{-2}] = [L^2 T^{-2}] = [LT^{-1}]^2$.$2$) $p + q - r = \beta$ is correct as $[LT^{-2}] [MLT^{-1}] / [MLT^{-2}] = [LT^{-1}]$.Thus, options $(A)$ and $(B)$ are correct.

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