The elastic behaviour of a material for linea | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) From the graph, the slope represents the ratio of strain to stress:$	ext{Slope} = \frac{	ext{strain}}{	ext{stress}} = \frac{4 	imes 10^{-10} \

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) From the graph, the slope represents the ratio of strain to stress:$	ext{Slope} = \frac{	ext{strain}}{	ext{stress}} = \frac{4 	imes 10^{-10} \; 	ext{unitless}}{80 \; N/m^2} = 0.05 	imes 10^{-10} \; m^2/N = 5 	imes 10^{-12} \; m^2/N$.Young's modulus $Y$ is the reciprocal of this slope:$Y = \frac{1}{	ext{Slope}} = \frac{1}{5 	imes 10^{-12}} = 0.2 	imes 10^{12} = 2 	imes 10^{11} \; N/m^2$.The energy density $u$ for a given strain $\epsilon$ is given by:$u = \frac{1}{2} Y \epsilon^2$.Substituting the values $Y = 2 	imes 10^{11} \; N/m^2$ and $\epsilon = 5 	imes 10^{-4}$:$u = \frac{1}{2} 	imes (2 	imes 10^{11}) 	imes (5 	imes 10^{-4})^2$$u = 10^{11} 	imes 25 	imes 10^{-8} = 25 	imes 10^3 \; J/m^3 = 25 \; kJ/m^3$.Thus, the correct option is $(C)$.

The elastic behaviour of a material for linear stress and linear strain is shown in the figure. The energy density for a linear strain of $5 \times 10^{-4}$ is $\dots \; kJ/m^3$. Assume that the material is elastic up to the linear strain of $5 \times 10^{-4}$.

Similar Questions

What is Elastic Moduli?

The breaking stress of aluminium is $7.5 \times 10^7 \,N m^{-2}$. The greatest length of aluminium wire that can hang vertically without breaking is ......... $\times 10^3 \,m$ (Density of aluminium is $2.7 \times 10^3 \,kg m^{-3}$ and $g = 9.8 \,m s^{-2}$)

Why do rail tracks have an '$I$-shaped' girder?

$A$ and $B$ are two wires. The radius of $A$ is twice that of $B$. They are stretched by the same load. Then the stress on $B$ is

Define shearing stress and write its formula.

Vedclass Test Series

Exam Paper Generator

Online Exam Module