Question

If for material Young's modulus is 6.6 x 1010 N/m2 and bulk
modulus is 11 x 101°N/m2 then the Poisson's ratio is
(A) 0.4
(B) 0.6
(C) 1
(D) infinity​

Answer 1

Given:

Young's modulus (Y) = $\sf 6.6 \times 10^{10} \ N/m^2$

Bulk modulus (B) = $\sf 11 \times 10^{10} \ N/m^2$

To Find:

Poisson's Ratio (σ)

Answer:

Relation between Young's modulus (Y), Bulk modulus (B) & Poisson's Ratio (σ):

$\boxed{ \boxed{ \bf{Y = 3B[1 - 2\sigma]}}}$

By substituting values we get:

$\rm \leadsto Y = 3B[1 - 2\sigma] \\ \\ \rm \leadsto 6.6 \times {10}^{10} = 3(11 \times {10}^{10}) [1 - 2\sigma] \\ \\ \rm \leadsto 6.6 \times \cancel{{10}^{10} } = 33 \times \cancel{{10}^{10}} [1 - 2\sigma] \\ \\ \rm \leadsto 1 - 2\sigma = \dfrac{6.6}{33} \\ \\ \rm \leadsto 1 - 2\sigma = 0.2 \\ \\\rm \leadsto 2\sigma = 1 - 0.2 \\ \\ \rm \leadsto 2\sigma =0.8 \\ \\ \rm \leadsto \sigma = \dfrac{0.8}{2} \\ \\ \rm \leadsto \sigma = 0.4$

$\therefore$ Poisson's Ratio (σ) = 0.4

Correct Option: $\mathfrak{(A) \ 0.4}$

Answer 2

σ = 3B - Y/6B

= 3 × 11 ×10^10 - 6.6 × 10^10/6 × 11 × 10^10

= 0.4