A steel $\operatorname{rod}\left(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\right.$ and $\left.\alpha=10^{-50}{ }^{\circ} \mathrm{C}^{-1}\right)$ of length 1 m and area of cross-section $1 \mathrm{~cm}^2$ is heated from $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$, without being allowed to extend or bend. What is the tension produced in the rod?
$$\begin{aligned} & \text { Given, Young's modulus of steel } Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \\ & \text { Coefficient of thermal expansion } \alpha=10^{-5}{ }^{\circ} \mathrm{C}^{-1} \\ & \text { Length } l=1 \mathrm{~m} \\ & \begin{aligned} & \text { Area of cross-section } A=1 \mathrm{~cm}^2=1 \times 10^{-4} \mathrm{~m}^2 \\ & \text { Increase in temperature } \Delta t=200^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}=200^{\circ} \mathrm{C} \\ & \text { Tension produced in steel rod }(F)=Y A \alpha \Delta t \\ &=2.0 \times 10^{11} \times 1 \times 10^{-4} \times 10^{-5} \times 200 \\ &=4 \times 10^4 \mathrm{~N} \end{aligned} \end{aligned}$$
To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1 \%$. (The bulk modulus of rubber is $9.8 \times 10^8 \mathrm{~N} / \mathrm{m}^2$; and the density of sea water is $10^3 \mathrm{~kg} / \mathrm{m}^3$ )
Given, Bulk modulus of rubber $(K)=9.8 \times 10^8 \mathrm{~N} / \mathrm{m}^2$
Density of sea water $(\rho)=10^3 \mathrm{~kg} / \mathrm{m}^3$
Percentage decrease in volume,
$\left(\frac{\Delta V}{V} \times 100\right)=0.1 \Rightarrow \frac{\Delta V}{V}=\frac{0.1}{100}$
$$\Rightarrow \quad \frac{\Delta V}{V}=\frac{1}{1000}$$
Let the rubber ball be taken upto depth $h$.
$\therefore \quad$ Change in pressure $(p)=h \rho g$
$\therefore \quad$ Bulk modulus $(K)=\left|\frac{p}{\Delta V / V}\right|=\frac{h \rho g}{(\Delta V / V)}$
$$\Rightarrow \quad h=\frac{K \times(\Delta V / V)}{\rho g}=\frac{9.8 \times 10^8 \times \frac{1}{1000}}{10^3 \times 9.8}=100 \mathrm{~m}$$
A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm . When the car just begins to move, the tension in the cable is 800 N . How much has the cable stretched? (Young's modulus for steel is $2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ )
$$ \begin{aligned} \text { Length of steel cable } l & =9.1 \mathrm{~m} \\ \text { Radius } r & =5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m} \\ \text { Tension in the cable } F & =800 \mathrm{~N} \\ \text { Young's modulus for steel } Y & =2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \\ \text { Change in length } \Delta l & =? \\ \text { Young's modulus }(Y) & =\frac{F}{A} \times \frac{l}{\Delta l} \Rightarrow \Delta l=\frac{F}{\pi r^2} \times \frac{l}{Y} \\ & =\frac{800}{3.14 \times\left(5 \times 10^{-3}\right)^2} \times \frac{9.1}{2 \times 10^{11}} \\ & =4.64 \times 10^{-4} \mathrm{~m} \end{aligned} $$
Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?
Since, ivory ball is more elastic than wet-clay ball, therefore, ivory ball tries to regain its original shape quickly. Hence, more energy and momentum is transferred to the ivory ball in comparison to the wet clay ball and therefore, ivory ball will rise higher after striking the floor.
Consider a long steel bar under a tensile stress due to forces $F$ acting at the edges along the length of the bar (figure). Consider a plane making an angle $\theta$ with the length. What are the tensile and shearing stresses on this plane?
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?
Consider the adjacent diagram.
Let the cross-sectional area of the bar be $A$. Consider the equilibrium of the plane aa'. A force $F$ must be acting on this plane making an angle $\frac{\pi}{2}-\theta$ with the normal $O N$. Resolving $F$ into components, along the plane (FP) and normal to the plane.
$$\begin{aligned} & F_P=F \cos \theta \\ & F_N=F \sin \theta \end{aligned}$$
$$\begin{aligned} &\text { Let the area of the face aa' be } A^{\prime} \text {, then }\\ &\begin{aligned} \frac{A}{A^{\prime}} & =\sin \theta \\ \therefore \quad A^{\prime} & =\frac{A}{\sin \theta} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { The tensile stress } & =\frac{\text { Normal force }}{\text { Area }}=\frac{F \sin \theta}{A^{\prime}} \\ & =\frac{F \sin \theta}{A / \sin \theta}=\frac{F}{A} \sin ^2 \theta \\ \text { Shearing stress } & =\frac{\text { Parallel force }}{\text { Area }} \\ & =\frac{F \cos \theta}{A / \sin \theta}=\frac{F}{A} \sin \theta \cdot \cos \theta \\ & =\frac{F}{2 A}(2 \sin \theta \cdot \cos \theta)=\frac{F}{2 A} \sin 2 \theta \end{aligned}$$
$$\begin{aligned} &\text { (a) For tensile stress to be maximum, } \sin ^2 \theta=1\\ &\begin{array}{lrl} \Rightarrow & \sin \theta & =1 \\ \Rightarrow & \theta & =\frac{\pi}{2} \end{array} \end{aligned}$$
$$\begin{aligned} &\text { (b) For shearing stress to be maximum, }\\ &\begin{aligned} & \Rightarrow \quad \sin 2 \theta =1 \\ & \Rightarrow \quad 2 \theta =\frac{\pi}{2} \\ & \Rightarrow\quad \theta =\frac{\pi}{4} \end{aligned} \end{aligned}$$