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.
$$\begin{aligned} & F_P=F \cos \theta \\ & F_N=F \sin \theta \end{aligned}$$
$$\begin{aligned} &\text { Let the area of the face aa' be } A^{\prime} \text {, then }\\ &\begin{aligned} \frac{A}{A^{\prime}} & =\sin \theta \\ \therefore \quad A^{\prime} & =\frac{A}{\sin \theta} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { The tensile stress } & =\frac{\text { Normal force }}{\text { Area }}=\frac{F \sin \theta}{A^{\prime}} \\ & =\frac{F \sin \theta}{A / \sin \theta}=\frac{F}{A} \sin ^2 \theta \\ \text { Shearing stress } & =\frac{\text { Parallel force }}{\text { Area }} \\ & =\frac{F \cos \theta}{A / \sin \theta}=\frac{F}{A} \sin \theta \cdot \cos \theta \\ & =\frac{F}{2 A}(2 \sin \theta \cdot \cos \theta)=\frac{F}{2 A} \sin 2 \theta \end{aligned}$$
$$\begin{aligned} &\text { (a) For tensile stress to be maximum, } \sin ^2 \theta=1\\ &\begin{array}{lrl} \Rightarrow & \sin \theta & =1 \\ \Rightarrow & \theta & =\frac{\pi}{2} \end{array} \end{aligned}$$
$$\begin{aligned} &\text { (b) For shearing stress to be maximum, }\\ &\begin{aligned} & \Rightarrow \quad \sin 2 \theta =1 \\ & \Rightarrow \quad 2 \theta =\frac{\pi}{2} \\ & \Rightarrow\quad \theta =\frac{\pi}{4} \end{aligned} \end{aligned}$$
(a) A steel wire of mass $\mu$ per unit length with a circular cross-section has a radius of 0.1 cm . The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming, the wire to be uniform and lateral strains $<<$ longitudinal strains, find the extension in the length of the wire. The density of steel is $7860 \mathrm{~kg} \mathrm{~m}^{-3}$. (Young's modulus $Y=2 \times 10^{11} \mathrm{Nm}^{-2}$.
(b) If the yield strength of steel is $2.5 \times 10^8 \mathrm{Nm}^{-2}$, what is the maximum weight that can be hung at the lower end of the wire?
Consider the diagram when a small element of length $d x$ is considered at $x$ from the load $(x=0)$
(a) Let $T(x)$ and $T(x+d x)$ are tensions on the two cross-sections a distance $d x$ apart, then -$-T(x+d x)+T(x)=d m g=\mu d x g$ (where $\mu$ is the mass/length). $(\because d m=\mu d x)$
$$d T=\mu g d x \quad[\because d T=T(x+d x)-T(x)]$$
$$\Rightarrow \quad T(x)=\mu g x+C \quad \text { (on integrating) }$$
$$\begin{aligned} &\begin{aligned} \end{aligned}\\ &\text { At } x=0, T(0)=M g \quad \Rightarrow \quad C=M g\\ &\begin{array}{ll} \therefore & T(x)=\mu g x+M g \end{array} \end{aligned}$$
$$\begin{aligned} &\text { Let the length } d x \text { at } x \text { increase by } d r \text {, then }\\ &\begin{aligned} \text { Young's modulus } Y & =\frac{\text { Stress }}{\text { Strain }} \\ \frac{T(x) / A}{d r / d x} & =Y \end{aligned} \end{aligned}$$
$$\begin{array}{ll} \Rightarrow & \frac{d r}{d x}=\frac{1}{Y A} T(x) \\ \Rightarrow & r=\frac{1}{Y A} \int_0^L(\mu g x+M g) d x \end{array}$$
$$\begin{aligned} & =\frac{1}{Y A}\left[\frac{\mu g x^2}{2}+M g x\right]_0^L \\ & =\frac{1}{Y A}\left[\frac{m g L^2}{2}+M g L\right]\quad \text{(m is the mass of the wire)} \end{aligned}$$
$$\begin{aligned} & A=\pi \times\left(10^{-3}\right)^2 \mathrm{~m}^2 \\ & Y=200 \times 10^9 \mathrm{Nm}^{-2} \\ & m=\pi \times\left(10^{-3}\right)^2 \times 10 \times 7860 \mathrm{~kg} \end{aligned}$$
$$\begin{aligned} \therefore \quad r & =\frac{1}{2 \times 10^{11} \times \pi \times 10^{-6}} \quad\left[\frac{\pi \times 786 \times 10^{-3} \times 10 \times 10}{2}+25 \times 10 \times 10\right] \\ & =\left[196.5 \times 10^{-6}+3.98 \times 10^{-3}\right] \approx 4 \times 10^{-3} \mathrm{~m} \end{aligned}$$
$$\begin{aligned} &\text { (b) Clearly tension will be maximum at } x=L\\ &\therefore \quad T=\mu g L+M g=(m+M) g \quad[\because m=\mu L] \end{aligned}$$
$$ \text { The yield force }=\left(\text { Yield strength } Y \text { ) area }=250 \times 10^6 \times \pi \times\left(10^{-3}\right)^2=250 \times \pi \mathrm{N}\right.$$
At yield point $$\quad$$ T = Yield force
$$\begin{array}{lrl} \Rightarrow & (m+M) g & =250 \times \pi \\ \therefore & m & =\pi \times\left(10^{-3}\right)^2 \times 10 \times 7860<< M \\ \text { Hence, } & M g & \approx 250 \times \pi \\ & M & =\frac{250 \times \pi}{10}=25 \times \pi \approx 75 \mathrm{~kg} . \end{array}$$