The elastic behavior of material for linear stress and linear strain is shown in the figure. The energy density for a linear strain of 5 × 10^-4 is _________ kJ/m³. Assume that material is elastic up to the linear strain of 5 × 10^-4.

Concept: 

In this question, we have to use the concept of stress and strain which can be defined by using Hooke's law, which states that stress is directly proportional to strain. 

stress ∝ strain     -----(1)

Here proportionality constant is 'λ' which is known as the elasticity constant. which can be defined as the ratio of stress and strain. 

λ = stress/ strain    -----(2)

Energy density can be defined as = (1/2) × stress × strain     -----(3)

Using equations (2) and (3) we get:

Energy density = (1/2) × λ × (strain)²     -----(4)

Calculations:

Given:

Linear strain = 5 × 10^-4 then energy density =?

The slope of the graph is = \(\frac{Y_{2}- Y_{1}}{X_{2} - X_{1}}\) = 1/λ    -----(5)

Slope = 10^-10/ 20 = 1/λ ⇒ λ = 20 × 10¹⁰ 

Using Equation (4) we get: 

Energy density = (1/2) × λ × (strain)2 

= (1/2) × 20× 10¹⁰ × (5× 10^-4)² = 25 × 10³ J/m³ = 25 KJ/m³

Hence, the correct answer is 25.