JEE Advanced · Physics · 2. Units & Dimensions
A dimensionless quantity is constructed in terms of electronic charge \(e\), permittivity of free space \(\varepsilon_0\), Planck's constant \(h\) and speed of light \(c\). If the dimensionless quantity is written as \(e^\alpha \varepsilon_0^\beta h^\gamma c^\delta\) and \(n\) is a non-zero integer, then \((\alpha, \beta, \gamma, \delta)\) is given by
- A \((2 n,-n,-n,-n)\)
- B \((n,-n,-2 n,-n)\)
- C \((n,-n,-n,-2 n)\)
- D \((2 n,-n,-2 n,-2 n)\)
Answer & Solution
Correct Answer
(A) \((2 n,-n,-n,-n)\)
Step-by-step Solution
Detailed explanation
For the quantity to be dimensionless
\(\mathrm{e}^\alpha \varepsilon_0^\beta \mathrm{h}^\gamma \mathrm{c}^{\mathrm{d}}=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0 \)
\( \Rightarrow(\mathrm{AT})^\alpha\left(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right)^\beta\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)^\gamma\)\(\left(\mathrm{LT}^{-1}\right)^\delta=\mathrm{A}^0 \mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \)
\( \therefore \alpha+2 \beta=0, \alpha+4 \beta-\gamma-\delta=0,-\beta+\gamma\) \(=0 ~\&~-3 \beta+2 \gamma+\delta=0 \)
\( \therefore \alpha=-2 \beta, \beta=\gamma ~\&~ \gamma=\delta\)
\(\therefore\) Option (1) satisfies the given condition
\(\mathrm{e}^\alpha \varepsilon_0^\beta \mathrm{h}^\gamma \mathrm{c}^{\mathrm{d}}=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0 \)
\( \Rightarrow(\mathrm{AT})^\alpha\left(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right)^\beta\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)^\gamma\)\(\left(\mathrm{LT}^{-1}\right)^\delta=\mathrm{A}^0 \mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \)
\( \therefore \alpha+2 \beta=0, \alpha+4 \beta-\gamma-\delta=0,-\beta+\gamma\) \(=0 ~\&~-3 \beta+2 \gamma+\delta=0 \)
\( \therefore \alpha=-2 \beta, \beta=\gamma ~\&~ \gamma=\delta\)
\(\therefore\) Option (1) satisfies the given condition
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