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MCQEasyJEE 2023Young's Modulus, Bulk & Rigidity Modulus

JEE Physics 2023 Question with Solution

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

  • A

    σ=3K−2η6K+2η\sigma = \frac{3K - 2\eta}{6K + 2\eta}

  • B

    σ=6K+2η3K−2η\sigma = \frac{6K + 2\eta}{3K - 2\eta}

  • C

    σ=3K+2η6K+2η\sigma = \frac{3K + 2\eta}{6K + 2\eta}

  • D

    σ=6K−2η3K+2η\sigma = \frac{6K - 2\eta}{3K + 2\eta}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Relations between Young's modulus EE, bulk modulus BB and modulus of rigidity GG are provided through standard elasticity formulas.

Find: The correct expression for Poisson's ratio σ\sigma in terms of bulk modulus and modulus of rigidity.

Using the given equations:

E=2G(1+σ)E = 2G(1 + \sigma)

and

E=3B(1−2σ)E = 3B(1 - 2\sigma)

Equating the two expressions for EE:

2G(1+σ)=3B(1−2σ)2G(1 + \sigma) = 3B(1 - 2\sigma)

Expanding both sides:

2G+2Gσ=3B−6Bσ2G + 2G\sigma = 3B - 6B\sigma

Collecting terms involving σ\sigma:

3B−2G=σ(2G+6B)3B - 2G = \sigma(2G + 6B)

Solving for σ\sigma:

σ=3B−2G2G+6B\sigma = \frac{3B - 2G}{2G + 6B}

Identifying B≡KB \equiv K and G≡ηG \equiv \eta, we get:

σ=3K−2η6K+2η\sigma = \frac{3K - 2\eta}{6K + 2\eta}

Therefore, the correct option is A.

Derivation of Poisson Ratio

Given:

  1. E=2G(1+σ)E = 2G(1 + \sigma)
  2. E=3B(1−2σ)E = 3B(1 - 2\sigma)

Find: Derive σ\sigma in terms of BB and GG, then match with the options.

From both expressions of EE:

2G(1+σ)=3B(1−2σ)2G(1 + \sigma) = 3B(1 - 2\sigma)

Now expand:

2G+2Gσ=3B−6Bσ2G + 2G\sigma = 3B - 6B\sigma

Bring constant terms to one side and σ\sigma terms to the other side:

3B−2G=2Gσ+6Bσ3B - 2G = 2G\sigma + 6B\sigma

Factor out σ\sigma:

3B−2G=σ(2G+6B)3B - 2G = \sigma(2G + 6B)

Hence,

σ=3B−2G2G+6B\sigma = \frac{3B - 2G}{2G + 6B}

Replacing BB by KK and GG by η\eta:

σ=3K−2η6K+2η\sigma = \frac{3K - 2\eta}{6K + 2\eta}

Thus, the correct answer is (A).

Common mistakes

  • Using the wrong elasticity relation, such as confusing bulk modulus with Young's modulus, gives an incorrect formula. Use E=2G(1+σ)E = 2G(1+\sigma) and E=3B(1−2σ)E = 3B(1-2\sigma) consistently.

  • Not recognizing that BB in the derivation is the same as KK and GG is the same as η\eta can lead to rejecting the correct option. Match the notation before comparing with the choices.

  • Making an algebraic sign error while moving the σ\sigma terms, especially in 3B−2G=σ(2G+6B)3B - 2G = \sigma(2G + 6B), changes the numerator or denominator. Rearrange carefully before solving.

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