JEE advance Physics PYQ series

Dimensionless Quantity in Physics

A dimensionless quantity is constructed in terms of electronic charge \( e \), permittivity of free space \( \varepsilon_0 \), Planck’s constant \( h \), and the 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:

Select the correct option:

• (A) \( (2n, -n, -n, -n) \)
• (B) \( (n, -n, -2n, -n) \)
• (C) \( (n, -n, -n, -2n) \)
• (D) \( (2n, -n, -2n, -2n) \)

The correct answer is (A) \( (2n, -n, -n, -n) \).

For the quantity to be dimensionless:

For the quantity to be dimensionless:

$$ e^{\alpha} \varepsilon_0^{\beta} h^{\gamma} c^{\delta} = M^0L^0T^0A^0 $$

Equating the dimensions:

$$ (AT)^{\alpha} \left(M^{-1}L^{-3}T^{4}A^{2}\right)^{\beta} \left(M^1L^2T^{-1}\right)^{\gamma} \left(LT^{-1}\right)^{\delta} = A^0M^0L^0T^0 $$

Solving for \( \alpha, \beta, \gamma, \delta \):

• \( \alpha = 2\beta \)
• \( \beta = \gamma \)
• \( \gamma = \delta \)

Equating the dimensions and solving gives us the conditions that match option (A).

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