Question

A tensile force of $2\times {10}^{3}N$ doubles the length of a rubber band pf cross-sectional area $2\times {10}^{-4}{m}^2$. The Young's modulus of elasticity of the rubber band is

• A. $4\times {10}^7{Nm}^2$
• B. $2\times {10}^7{Nm}^2$
• C. ${10}^7{Nm}^2$
• D. $0.5\times {10}^7{Nm}^2$

Answer 1

Answer: (C)

${10}^7{Nm}^2$

$\begin{array}{c}\text{Grven Tensile Force}=2\times {10}^{3}N\\ \text{cross sectponarea}=2\times {10}^{-4}{m}^2\end{array}$

$\begin{array}{c}\text{Let}l\text{be the length of wre initrally given that the length}\\ \text{becomes double after applying force. that means final length}\\ \text{will be}2l\text{.}\end{array}$

$\begin{array}{c}\text{Now, elongation}\begin{array}{cc}\Delta l& =2l-l\\ & =l\end{array}\\ \begin{array}{ccc}\text{Youngs modulus}y=\frac{\text{stress}}{\text{Strain}}& & \\ =& \frac{\text{Force}}{\text{Area}}/\left(\frac{\Delta l}{\ell }\right)& (\Delta l=l)\\ & =\frac{2X{10}^{3}N}{2\times {10}^{-4}{m}^2}\times \frac{l}{l}& \end{array}\end{array}$

$Y={10}^{7}N/m^2$