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JEE Advance 2020
paper 1
Multiple correct answers - Select all that apply
Question
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 ] $=\left[X^{\mathrm{P}}\right]$; $[$ linear momentum $]=\left[X^{\mathrm{q}}\right] ;[$ force $]=\left[X^{\mathrm{r}}\right]$. Then -
Select ALL correct options:
A
$\alpha+p=2 \beta$
B
$p+q-r=\beta$
C
$p-q+r=\alpha$
D
$p+q+r=\beta$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
⚠ Partially correct. Some answers are missing.
Solution
Sol. Given $\mathrm{L}=\mathrm{x}^\alpha$ $$ \begin{aligned} & \mathrm{LT}^{-1}=\mathrm{x}^\beta \\ & \mathrm{LT}^{-2}=\mathrm{x}^{\mathrm{p}} \\ & \mathrm{MLT}^{-1}=\mathrm{x}^{\mathrm{q}} \\ & \mathrm{MLT}^{-2}=\mathrm{x}^{\mathrm{r}} \\ & \quad \frac{(1)}{(2)} \Rightarrow \mathrm{T}=\mathrm{x}^{\alpha-\beta} \end{aligned} $$ From (3) $$ \begin{aligned} & \frac{\mathrm{x}^\alpha}{\mathrm{x}^{2(\alpha-\beta)}}=\mathrm{x}^{\mathrm{p}} \\ \Rightarrow & \alpha+\mathrm{p}=2 \beta \end{aligned} $$ From (4) $$ M=x^{q-\beta} $$ From (5) $\Rightarrow x^q=x^r x^{\alpha-\beta}$ $$ \Rightarrow \alpha+r-q=\beta $$ Replacing value ' $\alpha$ ' in equation (6) from (A) $$ \begin{aligned} & 2 \beta-\mathrm{p}+\mathrm{r}-\mathrm{q}=\beta \\ \Rightarrow & \mathrm{p}+\mathrm{q}-\mathrm{r}=\beta \end{aligned} $$ Replacing value of ' $\beta$ ' in equation (6) from (A) $$ \begin{aligned} & 2 \alpha+2 r-2 q=\alpha+p \\ & \alpha=p+2 q-2 r \end{aligned} $$
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