Question

A rod $100cm$ long and of $2cm\times 2cm$ cross-section is subjected to a pull of $1000kg$ force. The elasticity of the material is $2.0\times {{10}^{6}}kgc{{m}^{-2}}$. If the elongation of the rod is $x$ mm, find the value of $40x$.

Answer 1

Hint: Young’s modulus or modulus of elasticity of a material is equal to the ratio of stress on the material and strain of the material. Stress on a material is nothing but the force acting per unit area of the material. Strain refers to the deformation caused on the material, with the application of force.

Formula used:
$1)\sigma =\dfrac{F}{A}$
$2)\varepsilon =\dfrac{\Delta L}{L}$
$3)E=\dfrac{FL}{A\Delta L}$

Complete step-by-step answer:
When force is applied to a material such as a metal rod, the metal rod experiences stress. Stress is defined as the force applied on the material per unit area of cross-section of the material. Mathematically, stress is given by
$\sigma =\dfrac{F}{A}$
where
$\sigma $ is the stress on a material
$F$ is the force acting on the material
$A$ is the area of cross-section of the material
Let this be equation 1.
When the metal rod experiences stress, the rod tends to deform in shape. This deformation is nothing but strain, which is defined as the ratio of change in length of the metal rod to the actual length of the metal rod. Mathematically, strain is given by
$\varepsilon =\dfrac{\Delta L}{L}$
where
$\varepsilon $ is the strain of a material
$\Delta L$ is the change in length of the material when force is applied
$L$ is the actual length of the material
Let this be equation 2.
Young’s modulus or modulus of elasticity is the property of a material, which describes the relation between stress on a material and strain of the material. Mathematically, modulus of elasticity is given by
$E=\dfrac{stress}{strain}=\dfrac{\sigma }{\varepsilon }=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}=\dfrac{FL}{A\Delta L}$
where
$E$ is the modulus of elasticity of a material
$\sigma $ is the stress on the material
$\varepsilon $ is the strain of the material
$F$ is the force acting on the material
$A$ is the area of cross-section of the material
$\Delta L$ is the change in length of the material when force is applied
$L$ is the actual length of the material
Let this be equation 3.
Coming to our question, we are provided with a rod of length $100cm$ long and area of cross-section $2cm\times 2cm$. It is said that a force of $1000kg$ is applied on this rod. Modulus of elasticity of the material of the rod is given as $2.0\times {{10}^{6}}kgc{{m}^{-2}}$. If the elongation of the rod is $xmm$, we are required to find the value of $40x$.
From this data, it is clear that
$\begin{align}
  & L=100cm \\
 & A=2cm\times 2cm=4c{{m}^{2}} \\
 & F=1000kg \\
 & E=2\times {{10}^{6}}kgc{{m}^{-2}} \\
\end{align}$
Substituting these values in equation 3, we have
$E=\dfrac{FL}{A\Delta L}\Rightarrow 2\times {{10}^{6}}kgc{{m}^{-2}}=\dfrac{1000kg\times 100cm}{4c{{m}^{2}}\times \Delta L}\Rightarrow \Delta L=\dfrac{{{10}^{5}}kgcm}{8\times {{10}^{6}}kg}=0.0125cm$
Therefore, the change in length or the elongation of the rod is equal to $0.0125cm$.
Now, this value of elongation of rod is equated to $xmm$, to determine the value of $40x$, as asked in the question. Clearly,
$\begin{align}
  & \Delta L=0.0125cm=0.125mm=x \\
 & 40x=0.125\times 40=5 \\
\end{align}$
Therefore, the value of $40x$ is equal to $5$.

Note: Students should not hurry while doing such problems. They should make sure that they are not missing the last sentence or the important part of the question. For example, in this solution, we are not just asked to find the elongation of the rod, but to find its value in $mm$. This value has to be further multiplied by $40$, to satisfy the question. Therefore, it is important to read and understand the question carefully.