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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 $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
$(A)$ $\alpha+p=2 \beta$
$(B)$ $p+q-r=\beta$
$(C)$ $p-q+r=\alpha$
$(D)$ $p+q+r=\beta$
$A,B$
$A,C$
$A,D$
$B,C$
Solution
Given $L =x^\alpha$ $. . . . . . (1)$
$LT ^{-1}=x^\beta$ $. . . . . . (2)$
$LT ^{-2}=x^{ p }$ $. . . . . . (3)$
$MLT ^{-1}=x^q$ $. . . . . . (4)$
$MLT ^{-2}=x^{ I }V$ $. . . . . . (5)$
$\quad \frac{(1)}{(2)} \Rightarrow T =x^{\alpha-\beta}$
From $(3)$
$\frac{ x ^\alpha}{ x ^{2(\alpha-\beta)}}= x ^{ p }$
$\Rightarrow \alpha+ p =2 \beta$
From $(4)$
$M=x^{q-\beta}$
From $(5)$ $\Rightarrow x ^{ q }= x ^{ T } x ^{\alpha-\beta}$
$\Rightarrow \alpha+ r – q =\beta$
Replacing value ' $\alpha$ ' in equation $(6)$ from $(A)$
$2 \beta- p + r – q =\beta$
$\Rightarrow p + q – r =\beta$
Replacing value of ' $\beta$ ' in equation $(6)$ from $(A)$
$2 \alpha+2 r-2 q=\alpha+p$
$\alpha=p+2 q-2 r$
