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JEE Advanced Physics Units and Dimensions 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

In a particular system of units, a physical quantity can be expressed in terms of the electric charge e , electron mass m e . Planck's constant h , and Coulomb's constant k = 1 4 π ε 0 , where ε 0  is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is B = e α m e β h γ k δ . The value of α + β + γ + δ  is _______.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Given here:  B = e α m e β h γ k δ The dimensions of  e = Current × time = I T  and  m e = M , Planck's constant  h = Energy Frequency = M L 2 T - 2 T - 1 = M L 2 T - 1  and  k = 1 ε 0 = M - 1 L - 3 T 4 I 2 - 1 = M L 3 T - 4 I - 2 Using Principle of homogeneity, M 1 T - 2 I - 1 = I T α M β M L 2 T - 1 γ M L 3 T - 4 I - 2 δ Comparing both sides, we get   β + γ + δ = 1             . . . i 2 γ + 3 δ = 0           . . . ii α - γ - 4 δ = - 2           . . . iii α - 2 δ = - 1           . . . iv On solving above four equations, we get α = 3 ,   β = - 4 ,   γ = 3  and  δ = 2 So,  α + β + γ + δ = 4

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