Question

Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope.The original length of the rope is 2 m and the area of cross-section is. If it extends in length by 1 cm, then Young's modulus of the material of the rope will be

• A. ${10}^5\text{N/}{\text{m}}^{\text{2}}$
• B. ${10}^6\text{N/}{\text{m}}^{\text{2}}$
• C. ${10}^7\text{N/}{\text{m}}^{\text{2}}$
• D. ${10}^8\text{N/}{\text{m}}^{\text{2}}$

Answer 1

The correct option is D ${10}^8\text{N/}{\text{m}}^{\text{2}}$

$\begin{array}{c}\text{Given Tension in rope}F=100N\\ \text{Length}=2m\\ \text{Area of cross section}=2{cm}^2=2\times {\left({10}^2\right)}^{-2}{m}^2\\ \text{Elongation}=1cm={10}^{-2}m\end{array}$

$\begin{array}{c}\text{we know that}\Delta l=\frac{Fl}{AY}\\ \begin{array}{c}{10}^{-2}=\frac{100\times 2}{2\times {\left({10}^{-2}\right)}^2\times Y}\\ Y=\frac{{10}^2\times {10}^2}{{10}^{-4}}\end{array}\\ Y={10}^{8}N/m^2\end{array}$