Question

On increasing the length by $0.5\;mm$ in a steel wire of length $2\;m$ and area of cross-section $2\;cm^{2}$, the force required is ($Y$ for steel $2.2\times 10^{11}N/m^{2}$)

& A.1.1\times {{10}^{5}}N \\\ & B.1.1\times {{10}^{4}}N \\\ & C.1.1\times {{10}^{3}}N \\\ & D.1.1\times {{10}^{2}}N \\\ \end{aligned}$$

Answer 1

The elastic modulus or the Young’s modulus of the material is the ratio of tensile or compressive stress to the longitudinal strain. Using the given data and the formula of Young's modulus, we can find the force applied on the wire.

Formula used: $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

Complete step by step answer:
We know that a bulk like aluminium and copper experience bulk modulus, which is related to the elasticity of the material. It is the measure of how strong any given substance is, when subjected to some force. We know that the elastic moduli or the Young’s modulus of the material is the ratio of tensile or compressive stress to the longitudinal strain. i.e. $Y=\dfrac{stress}{strain}$, where stress is the force per unit area i.e.$stress=\dfrac{force}{area}$ and strain is the ratio of change in size or shape to the original shape or size i.e. $strain=\dfrac{change\; in\; shape}{original\; in\;shape}$. Then $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$.Also, Young's modulus is a constant for any given material.
Here, it is given that the length of the steel wire is $\Delta L=0.5mm$ and $L=2\;m$ and young’s
modulus is $Y=2.2\times 10^{11}N/m^{2}$, also given that the area of cross-section $A=2\;cm^{2}$.
Then, we have, $F=\dfrac{Y\Delta L A}{L}$
Substituting the values, we have, $F=\dfrac{2.2\times 10^{11}\times 2\times 10^{-4}\times 0.5\times 10^{-3}}{2}=1.1\times 10^{4}N$

So, the correct answer is “Option B”.

Note: Any bulk material experiences bulk modulus, which is related to the elasticity of the material. This is a very easy sum, provided one knows the formula of young’s modulus. Here, the units of various parameters are different. For easy calculations it is suggested to convert all the values to meters. Also, here we have two values for length , the smallest of the given values is considered as the change in length.