Two wires A and B of the same cross-section a | vedclass

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

Question

(A) The formula for Young's modulus $Y$ is given by $Y = \frac{T L}{A \Delta L}$,where $T$ is the tension,$L$ is the initial length,$A$ is the cross-s

Vedclass · Accepted Answer

$\frac{L_A}{L_B} = \frac{10}{11}$

Vedclass · Answer

(A) The formula for Young's modulus $Y$ is given by $Y = \frac{T L}{A \Delta L}$,where $T$ is the tension,$L$ is the initial length,$A$ is the cross-sectional area,and $\Delta L$ is the elongation.Given that the cross-sectional area $A$ and tension $T$ are the same for both wires,we have:$Y_A = \frac{T L_A}{A \Delta L_A}$ and $Y_B = \frac{T L_B}{A \Delta L_B}$Taking the ratio of the two moduli:$\frac{Y_A}{Y_B} = \frac{L_A}{L_B} 	imes \frac{\Delta L_B}{\Delta L_A}$We are given $\Delta L_B = 2 \Delta L_A$,so $\frac{\Delta L_B}{\Delta L_A} = 2$.Substituting the given values $Y_A = 2 	imes 10^{11} \ Nm^{-2}$ and $Y_B = 1.1 	imes 10^{11} \ Nm^{-2}$:$\frac{2 	imes 10^{11}}{1.1 	imes 10^{11}} = \frac{L_A}{L_B} 	imes 2$$\frac{2}{1.1} = \frac{L_A}{L_B} 	imes 2$$\frac{1}{1.1} = \frac{L_A}{L_B}$$\frac{L_A}{L_B} = \frac{10}{11}$

Similar Questions

The ratio of Young's modulus of the material of two wires is $2 : 3.$ If the same stress is applied on both,then the ratio of elastic energy per unit volume will be

An iron rod of length $2\, m$ and cross-sectional area of $50\, mm^2$ is stretched by $0.5\, mm$ when a mass of $250\, kg$ is hung from its lower end. The Young's modulus of the iron rod is:

Which of the following graphs correctly depicts the spring constant $k$ versus the length $l$ of the spring?

Vedclass Test Series

Exam Paper Generator

Online Exam Module