Question

2. The modulus of rigidity and poisson's ratio of
the material of a wire are 2.87*10^10 Nm-2 and
0.379 respectively. Find the value of young's
modulus of the material of the wire. *​

Answer 1

Answer:

7.915 * 10^10 Pa

Explanation:

use relation

E=2G(1+v)

where,

E= Young's Modulus

G= Shear Modulus or Modulus of Rigidity

v= Poisson's Ratio

E= 2*(2.87*10^10)* (1+0.379)

E= 7.915* 10^10 Pa

Answer 2

Given :

Modulus of rigidity (n) = 2.87 × 10¹⁰ Nm⁻²

Poisson's ratio (σ) = 0.379

To Find :

The value of Young's modulus of the material of the wire.

Solution :

The relation between Young's modulus (Y), modulus of rigidity (n) and Poisson's ratio (σ) can be given as -

Young's modulus (Y) = 2n(1 + σ)

                                  = 2×2.87×10¹⁰ (1 + 0.379)

                                  = 5.74 × 10¹⁰ × 1.379

                                  = 7.9 × 10¹⁰ N/m²

Therefore, the value of Young's modulus of the material of the wire is 7.9 × 10¹⁰ N/m².