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