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Physics Question on Elastic and inelastic collisions

Choose the option showing the correct relation between Poisson's ratio (σ), Bulk modulus (B) and modulus of rigidity (G).

A

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

B

σ=6B+2G3B−2G\sigma=\frac{6B+2G}{3B-2G}

C

σ=9BG3B+G\sigma=\frac{9BG}{3B+G}

D

B=3σ−3G6σ+2GB=\frac{3\sigma-3G}{6\sigma+2G}

Answer

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

Explanation

Solution

E = 2G(1 + σ) ….(1)
E = 3B(1 – 2σ) ….(2)
1=2G3G(1+σ1−2σ)1=\frac{2G}{3G}(\frac{1+\sigma}{1-2\sigma})
⇒ 3B – 6Bσ = 2G + 2Gσ
⇒ 3B – 2G = σ (2G + 6B)
σ=(3B−2G2G+6B)\sigma=(\frac{3B-2G}{2G+6B})

So, the correct answer is (A) : σ=3B−2G2G+6B\sigma=\frac{3B-2G}{2G+6B}