A dimensionless quantity is constructed in terms of electronic charge e , permittivity of free space _0 , Planck's constant h and speed of light c . If the dimensionless quantity is written as e^α _0^β h^γ c^ and n is a

Correct answer: A. (2 n,-n,-n,-n) For the quantity to be dimensionless aligned & e^α _0^β h^γ c^ d =M^0 ~L^0 ~T^0 ~A^0 & (AT)^α(M^ -1 ~L^ -3 ~T^4 ~A^2)^β(ML^2 ~T^ -1 )^γ(LT^ -1 )^ =A^0 M^0 ~L^0 ~T^0 & α+2 β=0, α+4 β-γ- =0,-β+γ=0 ~&~-3 β+2 γ+ =0 & α=-2 β, β=γ ~&~ γ= aligned Option (1) satisfies the given condition

JEE Advanced 2024Paper 1

Exam: JEE Advanced
Year: 2024
Paper: Paper 1
Subject: Physics
Chapter: Units and Dimensions

JEE Advanced 2024 · Paper 1 · Physics · Units and Dimensions

A dimensionless quantity is constructed in terms of electronic...

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

Options

A $(2 n,-n,-n,-n)$
B $(n,-n,-2 n,-n)$
C $(n,-n,-n,-2 n)$
D $(2 n,-n,-2 n,-2 n)$

Correct answer

A. $(2 n,-n,-n,-n)$

Step-by-step solution

For the quantity to be dimensionless

\begin{aligned} & \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\end{aligned}

\therefore$ Option (1) satisfies the given condition

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