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Properties of Matter question

2023 · 30 Jan · Shift 1 · Q43
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Properties of Matter question

2023 · 30 Jan · Shift 1 · Q43

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Choose the correct relationship between Poisson ratio (σ)(\sigma)(σ), bulk modulus (K) and modulus of rigidity (η)(\eta)(η) of a given solid object :
  1. A
    σ=3K+2η6K+2η\sigma=\frac{3 K+2 \eta}{6 K+2 \eta}σ=6K+2η3K+2η​
  2. B
    σ=3K−2η6K+2η\sigma=\frac{3 K-2 \eta}{6 K+2 \eta}σ=6K+2η3K−2η​
  3. C
    σ=6K+2η3K−2η\sigma=\frac{6 K+2 \eta}{3 K-2 \eta}σ=3K−2η6K+2η​
  4. D
    σ=6K−2η3K−2η\sigma=\frac{6 K-2 \eta}{3 K-2 \eta}σ=3K−2η6K−2η​
View written solutionFree

Correct answer: B

  1. Use the standard elastic relation

For an isotropic solid, the relation between:

  • Bulk modulus KKK
  • Modulus of rigidity η\etaη (shear modulus GGG)
  • Poisson ratio σ\sigmaσ

is

K=2η(1+σ)3(1−2σ)K = \frac{2\eta(1+\sigma)}{3(1-2\sigma)}K=3(1−2σ)2η(1+σ)​
  1. Solve this equation for σ\sigmaσ

Starting with

K=2η(1+σ)3(1−2σ)K = \frac{2\eta(1+\sigma)}{3(1-2\sigma)}K=3(1−2σ)2η(1+σ)​

Cross-multiply:

3K(1−2σ)=2η(1+σ)3K(1-2\sigma)=2\eta(1+\sigma)3K(1−2σ)=2η(1+σ)

Expand both sides:

3K−6Kσ=2η+2ησ3K-6K\sigma = 2\eta + 2\eta\sigma3K−6Kσ=2η+2ησ

Bring terms containing σ\sigmaσ to one side:

3K−2η=6Kσ+2ησ3K - 2\eta = 6K\sigma + 2\eta\sigma3K−2η=6Kσ+2ησ

Factor out σ\sigmaσ:

3K−2η=σ(6K+2η)3K - 2\eta = \sigma(6K+2\eta)3K−2η=σ(6K+2η)

Hence,

σ=3K−2η6K+2η\sigma = \frac{3K-2\eta}{6K+2\eta}σ=6K+2η3K−2η​
  1. Match with the options

This corresponds to:

σ=3K−2η6K+2η\boxed{\sigma=\frac{3K-2\eta}{6K+2\eta}}σ=6K+2η3K−2η​​

So the correct option is B.

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