For a given material, the Young’s modulus is 2.4 times that of rigidity modulus. Its Poisson’s ratio is

The correct option is 0.2.

Step-by-Step Explanation:

To find the Poisson's ratio, we use the standard elastic relation that connects Young's modulus, the modulus of rigidity, and Poisson's ratio for an isotropic material.

The relation is given by the formula:

$Y=2\eta (1+\sigma )$

where:
• $Y$ is the Young's modulus of the material,
• $\eta$ is the rigidity modulus (shear modulus), and
• $\sigma$ is the Poisson's ratio.

According to the question, the Young's modulus is 2.4 times that of the rigidity modulus:

$Y=2.4\eta$

Now, substitute this value into the relation formula:

$2.4\eta =2\eta (1+\sigma )$

Since the rigidity modulus $\eta$ is a non-zero positive quantity, we can divide both sides of the equation by $\eta$:

$2.4=2(1+\sigma )$

Next, divide both sides by 2:

$1.2=1+\sigma$

Subtract 1 from both sides to solve for Poisson's ratio ($\sigma$):

$\sigma =1.2-1$

$\sigma =0.2$

Therefore, the Poisson's ratio of the material is 0.2.