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Consider a wire of length $l$ and cross-sectional area $A$ fixed at one end. When other end of the wire is pulled with a force $F$, the wire gets slightly elongated by $\Delta l$ and the external force gets balanced by the internal forces. The internal forces in a wire are along its length i.e., normal to cross-sectional plane. The internal force per unit area is called longitudinal stress
\begin{align}
\sigma=\frac{F}{A}
\end{align}
Since the wire is under tension, the stress is called tensile stress. The unit of stress is N/m^{2} or Pascal.

The increase in length per unit orginal length is called strain i.e., \begin{align} \text{Longitudinal strain}=\frac{\Delta l}{l} \end{align} The strain is a dimesnionless quantity. It is usually expressed as percentage increase in length.

If a rod is compressed then the stress in it is compressive stress, and the strain is compressive strain.

The tangential force per unit area is called **shear stress** i.e.,
\begin{align}
\sigma_s=\frac{F_s}{A}
\end{align}

The shearing strain is defined as i.e., \begin{align} \text{Shearing strain}=\frac{x}{l}=\tan\theta\approx\theta \end{align}

According to **Hooke's law**, the stress in a body is proportional to the corresponding strain. The constant of proportionality is called modulus of elasticity. Hooke's law is valid for small value of stress and for small deformations (strain).

The ratio of longitudinal stress to longitudinal strain, within the elastic limit to which Hooke's law is applicable, is called Young's modulus of the material i.e.,
\begin{align}
Y & =\frac{\text{longitudinal stress}}{\text{longitudinal strain}} \\
&=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}.
\end{align}
The unit of Young's modulus is same as that of stress i.e., N/m^{2}.

When uniform pressure $p$ is applied over the whole surface of a body, it produces a uniform compression. If the pressure is small, the compression is proportional to the pressure. The ratio of the pressure to the volume strain is called Bulk modulus of the material, \begin{align} B=-\frac{p}{\Delta V/V} \end{align}

The ratio of shearing (or tangential) stress to sharing strain is called shear modulus or modulus of rigidity i.e., \begin{align} \eta &=\frac{F_s/A}{x/l}=\frac{F_sl}{xA} \\ &=\frac{F_s}{A\theta} \end{align}

Consider pulling of a wire by a force $F$. The longitudinal dimension (length) of the wire is increased from $l$ to $l+\Delta l$. Its lateral dimension (diameter) is decreased from $D$ to $D-\Delta D$. Within elastic limits there is a complete proportionality between the lateral strain and the longitudinal strain and the ratio between the lateral strain and longitudinal strain is called **Poisson's ratio** i.e.,
\begin{align}
\sigma=\frac{\Delta D/D}{\Delta l/l}=\frac{l}{D}\frac{\Delta D}{\Delta l}
\end{align}
Poisson's ratio is a dimensional quentity.

Young's modulus, Bulk modulus and shear modulus are related by \begin{align} Y&=\frac{9\eta B}{3B+\eta} \\ \sigma&=\frac{3B-2\eta}{6B+2\eta}. \end{align}

Consider pulling of a wire. When the strain is small (say less than 1%), the stress is proportional to the strain. This is the region where Hooke's law is valid. The point on the stress-strain curve upto which stress is proportional to strain is called proportional limit. The body return to its original state after removal of external force.

If the strain is increased a little bit, the stress is not proportional to the strain. This region is called plastic region. The body does not return to its original state in the plastic region.

The fracture point is the point at which the wire break. The stress at fracture point is called breaking stress.

If large deformation takes place between the elastic limit and the fracture point, the material is called **ductile**. If it breaks soon after the elastic limit is crossed, it is called **brittle**.

For a linear spring described in textbooks, the extension is proportional to the applied force. But most materials are linear for a small extension range only. In this experiment you will study the relation between the load applied on a hanging bicycle valve tube and its extension. Also rubber shows hysteresis so the extension depends not only on the force but also on the history of stretching.

You need a bicycle valve tube with knots at the ends, a stand to suspend the tube, a weight hanger, weights, meter scale vertically fixed, graph paper etc.

Measure the natural length of the tube. Suspend the tube in the stand and the weight hanger from its lower end. Put the weights one by one and note the reading of appropriate point on the scale. Each time write the load and the scale reading in a table. Calculate the extension. Once you have gone up to say 1 kg, remove the weights one by one and each time note the extension. Write in the same table. Make sure you monotonically increase the load and also monotonically decrease the load. Also you have to give sufficient time after putting or removing the load for the tube to settle down.

Draw a graph with load on the \(x\)-axis and extension on the other axis. The area under the curve gives the work done by the load. Calculate the work done in the increasing load cycle and also in the decreasing load cycle. Your answer should be in joules. Is any of the two positive? Is any of the two negative? What is the total energy dissipated in the whole process?

**Problem (IIT JEE 1996):**
The extension in a string, obeying Hooke's law, is $x$. The speed of transverse wave in the stretched string is $v$. If the extension in the string is increased to $1.5 x$, the speed of transverse wave will be,

- $1.22 v$
- $0.61 v$
- $1.50 v$
- $0.75 v$

**Solution: **
The velocity of transverse wave in a string under tension $T$ and mass per unit length $\mu$ is given by,
\begin{align}
v=\sqrt{T/\mu}.
\end{align}
The tension in the string is related to its extension $x$ by Hooke's law,
\begin{align}
T=kx,
\end{align}
where $k$ is a constant. The tension is $kx$ in the first case and $1.5kx$ in the second case. Substitute $T=1.5kx$ in first equation to get velocity in the second case,
\begin{align}
v^\prime & =\sqrt{1.5kx/\mu} \\
&=\sqrt{1.5}\,\sqrt{kx/\mu} \\
&=1.22v.\nonumber
\end{align}
We have assumed $x$ to be much smaller than the string length to keep $\mu$ constant.