QJEEA-2024-01: JEE Advanced 2024 JEEA-2024-01 Solution: Q.1  A dimensionless quantity is constructed in terms …

Given Dimensional Formulas:

• e (charge): [IT]
• ε₀ (permittivity): [I²T⁴M⁻¹L⁻³]
• h (Planck's constant): [ML²T⁻¹]
• c (speed of light): [LT⁻¹]

Solution:

1. Set Up Dimensional Equation:

• e^αε₀^βh^γc^δ = [M⁰L⁰T⁰I⁰] (dimensionless)

1. Substitute Dimensional Formulas:

• [IT]^α[I²T⁴M⁻¹L⁻³]^β[ML²T⁻¹]^γ[LT⁻¹]^δ = [M⁰L⁰T⁰I⁰]

1. Equate Exponents for Each Dimension:

• Mass (M): -β + γ = 0 (1)
• Length (L): -3β + 2γ + δ = 0 (2)
• Time (T): α + 4β - γ - δ = 0 (3)
• Current (I): α + 2β = 0 (4)

1. Solve the Equations:

• From equation (4): α = -2β
• From equation (1): γ = β
• Substitute γ = β into equation (2): -3β + 2β + δ = 0, which simplifies to -β + δ = 0, so δ = β.
• Substitute γ = β and δ = β into equation (3): α + 4β - β - β = 0, which simplifies to α + 2β = 0. This confirms the solution.

1. Express in Terms of 'n':

• Let β = -n (where n is a non-zero integer).
• Then:
• α = -2β = 2n
• γ = β = -n
• δ = β = -n

1. Result:

• Therefore, (α, β, γ, δ) = (2n, -n, -n, -n).

Answer:

The exponents (α, β, γ, δ) are (2n, -n, -n, -n), which corresponds to option A.