Elasticity

Elasticity is a fundamental concept in physics and economics that measures how much an object or material can stretch or compress when subjected to external forces. In physics, it's about how materials like rubber bands, springs, or even metal wires respond to being pulled or pressed, returning to their original shape once the force is removed. Economically, elasticity helps us understand how changes in price can affect the demand or supply of a product. For instance, a rubber band stretching when pulled represents elastic behaviour, similar to how demand might increase when prices drop. In everyday life, elasticity is seen when we stretch a rubber band, squeeze a stress ball, or even when we observe how gasoline prices affect our driving habits. This concept helps bridge the gap between theoretical principles and practical experiences, making it easier to understand how things react and adapt to changes.

Elasticity

Elasticity refers to the ability of a material or substance to return to its original shape and size after being deformed by an external force. This property is crucial in both physical and economic contexts. In physics, elasticity describes how materials like rubber, metals, or polymers respond to stretching, compressing, or bending forces, and how they return to their initial state once the force is removed. For example, when you stretch a rubber band and then let it go, it snaps back to its original shape due to its elastic nature.

Reason of Elasticity

In solids, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to neighbouring molecules which keep molecules in the position of stable equilibrium. These forces are known as intermolecular forces. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic (or intermolecular) distances. When the deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size. The restoring mechanism can be visualised by taking a model of the spring-ball system. Here the balls represent atoms and springs represent interatomic forces. If you displace any ball from its equilibrium position, the spring system tries to restore the ball back to its original position.

Fig:- Spring-ball model for the illustration of elastic behaviour of solids.

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Solved Examples Based on Elasticity

Example 1: A brass rod of length 2 m and cross-sectional area $2.0 \mathrm{~cm}^2$ is attached end to end to a steel rod of length $L$ and cross-sectional area $1.0 \mathrm{~cm}^2$. The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^4 \mathrm{~N}$ at its ends. If the elongations of two rods are equal, the length of the steel $(L)$ is

Given,
$Y_{\text {Brass }}=1.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ and
$Y_{\text {steel }}=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

1) 1.5 m
2) 1.8 m
3) $1 / m$
4) 2 m

Solution:

$(\Delta l)_b=(\Delta l)_s$

$\begin{aligned} & \left(\frac{F l}{A Y}\right)_b=\left(\frac{F l}{A Y}\right)_s \quad\left(F_b=F_s\right) \\ & \left(\frac{l}{A Y}\right)_b=\left(\frac{l}{A Y}\right)_{R_b} \\ & l_s=\left(\frac{A_s Y_s}{A_b Y_b}\right) l_b \\ & l_s=\left(\frac{1.0 \times 2.0 \times 10^{11}}{2.0 \times 1.0 \times 10^{11}}\right)(2 \mathrm{~m}) \\ & l_s=2 \mathrm{~m}\end{aligned}$

Hence, the answer is the option (4).

Example 2: The Young's modulus of brass and steel are $1.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$ and $2 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$ respectively. A brass wire and a steel wire of the same length are extended by 1mm under the same force, the radii of brass and steel wires are R_B and R_S respectively. Then-

${ }^{1)} R_S=\sqrt{2} R_B$
2) $R_S=\frac{R_B}{\sqrt{2}}$
3) $R_S=4 R_B$
${ }^{4)} R_S=\frac{R_B}{4}$

Solution:

$
\Delta l=\frac{F_l}{A_y}=\frac{F_l}{\Pi R^2 Y}
$
$\Delta l, F$ and $l$ are same. Hence, $R^2=$ Constant
$
\begin{aligned}
& \frac{R_S}{R_B}=\sqrt{\frac{Y_B}{Y_S}}=\sqrt{\frac{1}{2}} \\
& R_S=\frac{R_B}{\sqrt{2}}
\end{aligned}
$

Hence, the answer is the option (2).

Example 3: A copper wire $y=10^{11} \mathrm{~N} / \mathrm{m}^2$ of length 8 m and a steel wire of $y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ length 4 m, each of 0.5 cm² cross -section are fastened end to end and stretched with a tension of 500 N.

1)Elongation in copper wire is 0.8 mm.

2)Elongation in steel is $\left(\frac{1}{4}\right) t h$ the elongation in copper wire.

3) Total elongation is 1.0 mm.

4)All of the above

Solution:

$(\Delta l)_c=\left(\frac{\mathrm{Fl}}{\mathrm{AY}}\right)_c=\frac{500 \times 8}{0.5 \times 10^{-4} \times 10^{11}}$

$\begin{aligned} & (\Delta l)_c=0.8 \times 10^{-3} \mathrm{~m}=0.8 \mathrm{~mm} \\ & \begin{aligned}(\Delta l)_S=\left(\frac{F l}{A Y}\right)_S \\ =\frac{500 \times 4}{0.5 \times 10^{-4} \times 2 \times 10^{11}}=0.2 \times 10^{-3} \mathrm{~m} \\ =0.2 \mathrm{~mm}\end{aligned} \\ & (\Delta l)_s=\frac{1}{4}(\Delta l)_c \\ & \Delta l=0.9+0.2=1.00 \mathrm{~mm}\end{aligned}$

Hence, the answer is the option (3).
Example 4: A uniform cylinder rod of length L, cross-sectional area A and Young's modulus y are acted upon by the forces shown in the figure. The elongation of the rod is

1) $\frac{3 F L}{5 A Y}$
2) $\frac{2 F L}{5 A Y}$
3) $\frac{3 F L}{8 \mathrm{AY}}$
4) $\frac{8 F L}{3 A Y}$

Solution:

The free-body diagrams of the two parts are

Both parts are stretched. Therefore, total elongation
$\begin{aligned} \Delta l & =\Delta l_1+\Delta l_2 \\ \Delta l & =\frac{3 F\left(\frac{2 L}{3}\right)}{A Y}+\frac{2 F\left(\frac{L}{3}\right)}{A Y} \\ \Delta l & =\frac{8 F L}{3 A Y}\end{aligned}$

Hence, the answer is the option (4).

Example 5: A rod of mass M, length l and cross-sectional area A, made of material of Young's modulus Y is rotated about its one end in the horizontal plane with constant angular speed $\omega$.
It's extension is

1) $\frac{M \omega^2 l^2}{3 A Y}$
2) $\frac{M \omega^2 l^2}{A Y}$
3) $\frac{M \omega^2 l^3}{A Y}$
4) $\frac{M \omega^2 l^2}{4 A Y}$

Solution:

For the element of mass dM

$
\begin{aligned}
& -(d F)=(d M) \omega^2 x \\
& F=-\int_l^x\left(\frac{M}{l} d x\right) \omega^2 x \\
& F=\frac{M}{2 l} \omega^2\left(l^2-x^2\right)
\end{aligned}
$

Now, using $d l=\frac{F(d x)}{A Y}$
$
\begin{aligned}
& d l=\frac{M \omega^2}{2 l} \frac{\left(l^2-x^2\right)}{A Y} d x \\
& \Delta l=\frac{M \omega^2}{2 Y A l} \int_0^l\left(l^2-x^2\right) d x
\end{aligned}
$
$
\Delta l=\frac{M \omega^2 l^2}{3 A Y}
$

Hence, the answer is the option (1).

Summary

Elasticity is the property of materials that allows them to return to their original shape after deformation. In physics, it's demonstrated by how materials like rubber bands or springs respond to forces, while in economics, it relates to how price changes affect supply and demand. The concept is explained through the behavior of atomic forces in solids and is quantified using Young's modulus. Practical examples, such as the elongation of rods and wires under tension, illustrate the application of elasticity principles in real-life scenarios.