Two wires of the same length and material are | vedclass

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

Question

(A) The elongation $\Delta L$ of a wire is given by the formula $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-s

Vedclass · Accepted Answer

$4:3$

Vedclass · Answer

(A) The elongation $\Delta L$ of a wire is given by the formula $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.Since $F$,$L$,and $Y$ are the same for both wires,$\Delta L \propto \frac{1}{A}$.The mass $m$ of a wire is given by $m = \rho AL$,where $\rho$ is the density. Since $\rho$ and $L$ are the same,$m \propto A$.Therefore,$\Delta L \propto \frac{1}{m}$.Given the ratio of masses $\frac{m_1}{m_2} = \frac{3}{4}$,the ratio of elongations is $\frac{\Delta L_1}{\Delta L_2} = \frac{m_2}{m_1} = \frac{4}{3}$.

Two wires of the same length and material are stretched by the same force. If their masses are in the ratio $3:4$,then the ratio of their elongations is

Similar Questions

If Young's modulus of iron is $2 \times 10^{11} \ N/m^2$ and the interatomic spacing between two molecules is $3 \times 10^{-10} \ m$,the interatomic force constant is ......... $N/m$.

If the ratio of lengths,radii,and Young's moduli of steel and brass wires in the figure are $a, b,$ and $c$ respectively,then the corresponding ratio of increase in their lengths is

When a force of $4 \, N$ is applied to a wire,its length is $a \, m$. When a force of $5 \, N$ is applied,its length is $b \, m$. What will be its length when a force of $9 \, N$ is applied?

If the density of the material increases,the value of Young's modulus

Two wires of the same length and same cross-sectional area are suspended as shown in the figure. Their Young's moduli are $Y_1$ and $Y_2$. What is their equivalent Young's modulus?

Vedclass Test Series

Exam Paper Generator

Online Exam Module