JEE Main 2023 · 30 January · morning · Single correct · +4 / −1

Choose the correct relationship between Poisson ratio (σ), bulk modulus (K) and modulus of rigidity (η) of a given solid object :

  1. (a) σ=(3 K+2 η)/(6 K+2 η)
  2. (b) σ=(3 K-2 η)/(6 K+2 η)
  3. (c) σ=(6 K+2 η)/(3 K-2 η)
  4. (d) σ=(6 K-2 η)/(3 K-2 η)
Answer and solution

Answer (b) σ=(3 K-2 η)/(6 K+2 η)

Solution

Physics sir · 4 steps

Write Young's modulus two ways, once with K and once with η, then set them equal and solve for σ.

  1. Two formulas for Y. Young's modulus links to the bulk modulus as Y = 3K(1-2σ) and to the modulus of rigidity as Y = 2η(1+σ). Both describe the same solid, so they must be equal.

    3K(1-2σ) = 2η(1+σ)

  2. Collect σ terms. Expanding gives 3K - 6Kσ = 2η + 2ησ. Put the σ terms on one side: 3K - 2η = σ(6K + 2η), so σ = (3K-2η)/(6K+2η).
  3. Quick sanity check. Poisson's ratio must lie between -1 and 0.5. Putting η = 0 gives 3K/6K = 0.5 and putting K = 0 gives -2η/2η = -1, exactly the two limits.
  4. So. The correct relation is σ = (3K-2η)/(6K+2η), which is option (b).

The idea to keep. To relate any three of Y, K, η and σ, write Y in terms of the other moduli, equate the expressions and solve. Then test the limits σ = 0.5 and σ = −1 to confirm.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 90 times in JEE Main since 2020. This question is from 2023.