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JEE Advanced Physics Units and Dimensions 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Physics Question (2023) — Solution

Question

Young’s modulus of elasticity Y  is expressed in terms of three derived quantities, namely, the gravitational constant G , Planck’s constant h and the speed of light c , as Y = c α h β G γ . Which of the following is the correct option?

Options

  1. A. α = 7 ,   β = - 1 ,   γ = – 2
  2. B. α = – 7 ,   β = – 1 ,   γ = – 2
  3. C. α = 7 ,   β = – 1 ,   γ = 2
  4. D. α = – 7 ,   β = 1 ,   γ = – 2

Answer

A. α = 7 ,   β = - 1 ,   γ = – 2

Step-by-step solution

Dimensions of Young's modulus:  M 1 L – 1 T   – 2 Dimensions of velocity of light:  M 0 L 1 T – 1 Dimensions of Planck's constant:    M 1 L 2   T – 1 Dimensions of Gravitational constant:  M – 1 L 3 T – 2 Now, as given in the question Y = c α h β G γ ⇒ M 1 L – 1 T   – 2 = M 0 L 1 T – 1 α   M 1 L 2   T – 1 β M – 1 L 3 T – 2 γ Comparing exponent of the M, L & T both sides, we get 1 = β - γ   ⇒   β = 1 + γ       . . . 1 - 1 = α + 2 β + 3 γ         . . . 2  and  - 2 = - α - β - 2 γ       . . . 3 From equation(1) and (2), we get - 1 = α + 2 1 + γ + 3 γ ⇒ α + 5 γ = - 3       . . . 4 From equation(1) and (3), we get - 2 = - α - 1 + γ - 2 γ ⇒ α + 3 γ = 1         . . . 5 From equation(4) and equation(5), we get α = 7 ,   γ = – 2  and then  β = 1 + γ = - 1 .

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