One end of a steel wire of radius r is fixed | vedclass

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

Question

(B) The longitudinal strain $\epsilon$ is given by $\epsilon = \frac{	ext{Stress}}{Y} = \frac{F}{A \cdot Y}$,where $F$ is the tension,$A$ is the cros

Vedclass · Accepted Answer

$1: 6$

Vedclass · Answer

(B) The longitudinal strain $\epsilon$ is given by $\epsilon = \frac{	ext{Stress}}{Y} = \frac{F}{A \cdot Y}$,where $F$ is the tension,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.For the steel wire: Tension $F_s = (3 + 2) \ kg 	imes g = 5g$. Area $A_s = \pi r^2$. Young's modulus $Y_s = 20 	imes 10^{10} \ Nm^{-2}$.Strain in steel $\epsilon_s = \frac{5g}{\pi r^2 \cdot 20 	imes 10^{10}}$.For the copper wire: Tension $F_c = 2 \ kg 	imes g = 2g$. Area $A_c = \pi (2r)^2 = 4\pi r^2$. Young's modulus $Y_c = 12 	imes 10^{10} \ Nm^{-2}$.Strain in copper $\epsilon_c = \frac{2g}{4\pi r^2 \cdot 12 	imes 10^{10}} = \frac{g}{2\pi r^2 \cdot 12 	imes 10^{10}} = \frac{g}{24\pi r^2 	imes 10^{10}}$.Ratio $\frac{\epsilon_c}{\epsilon_s} = \frac{g}{24\pi r^2 	imes 10^{10}} 	imes \frac{20\pi r^2 	imes 10^{10}}{5g} = \frac{20}{24 	imes 5} = \frac{20}{120} = \frac{1}{6}$.Thus,the ratio is $1: 6$.

Similar Questions

The length of a light string is $1.4 \ m$ when the tension on it is $5 \ N$. If the tension increases to $7 \ N$,the length of the string is $1.56 \ m$. The original length of the string is . . . . . . $m$.

Write the unit and dimensional formula of modulus of elasticity.

The dimensions of four wires of the same material are given below. In which wire will the increase in length be maximum when the same tension is applied?

Vedclass Test Series

Exam Paper Generator

Online Exam Module