JEE Advanced · Physics · 2. Units & Dimensions
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 as follows: ; ;; . Then
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
Step-by-step Solution
Detailed explanation
\(\left[\frac{\text { momentum }}{\text { Force }}\right]= X ^{p-r}\)
\(\Rightarrow\left[\frac{m v}{m a}\right]=X^{p-r}\)
\(\Rightarrow\left[\frac{v}{a}\right]=X^{p-r}\)
but given that \(\frac{v}{a}=X^{\beta-p}\)
So,
\(p-r=\beta-p\)
\(\Rightarrow 2 p=\beta+r \quad \ldots(1)\)
\(\left[\frac{\text { speed }^2}{\text { acceleration }}\right]=[\) position \(]\)
\(\Rightarrow \frac{X^{2 \beta}}{X^p}=X^\alpha\)
\(\Rightarrow 2 \beta-p=\alpha\)
\(\Rightarrow p+\alpha=2 \beta\quad\ldots(2)\)
and
\(\left[\frac{\text { force }}{\text { linear momentum }}\right]=\left[\frac{\text { acceleration }}{\text { speed }}\right]\)
\(\Rightarrow\left[\frac{X^r}{X^q}\right]=\left[\frac{X^p}{X^\beta}\right]\)
\(\Rightarrow r-q=p-\beta\)
\(\Rightarrow p+q-r=\beta\quad\ldots(3)\)
\(\Rightarrow\left[\frac{m v}{m a}\right]=X^{p-r}\)
\(\Rightarrow\left[\frac{v}{a}\right]=X^{p-r}\)
but given that \(\frac{v}{a}=X^{\beta-p}\)
So,
\(p-r=\beta-p\)
\(\Rightarrow 2 p=\beta+r \quad \ldots(1)\)
\(\left[\frac{\text { speed }^2}{\text { acceleration }}\right]=[\) position \(]\)
\(\Rightarrow \frac{X^{2 \beta}}{X^p}=X^\alpha\)
\(\Rightarrow 2 \beta-p=\alpha\)
\(\Rightarrow p+\alpha=2 \beta\quad\ldots(2)\)
and
\(\left[\frac{\text { force }}{\text { linear momentum }}\right]=\left[\frac{\text { acceleration }}{\text { speed }}\right]\)
\(\Rightarrow\left[\frac{X^r}{X^q}\right]=\left[\frac{X^p}{X^\beta}\right]\)
\(\Rightarrow r-q=p-\beta\)
\(\Rightarrow p+q-r=\beta\quad\ldots(3)\)
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