What is the Bulk modulus for a perfect rigid body?
Bulk modulus $(K)=\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}$
For perfectly rigid body, change in volume $\Delta V=0$
$$\therefore\quad K=\frac{p V}{0}=\infty$$
Therefore, bulk modulus for a perfectly rigid body is infinity ( $\infty$ ).
A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of the same material of length $2 L$ and radius $2 r$, is pulled by a force $2 f$. Find the increase in length of this wire.
The situation is shown in the diagram.
Now, $$\quad$$ Young's modulus $(Y)=\frac{f}{A} \times \frac{L}{l}$
For first wire, $$Y=\frac{f}{\pi r^2} \times \frac{L}{l} \quad \text{... (i)}$$
For second wire,
$$\begin{aligned} Y & =\frac{2 f}{\pi(2 r)^2} \times \frac{2 L}{l^{\prime}} \\ & =\frac{f}{\pi r^2} \times \frac{L}{l^{\prime}}\quad \text{... (ii)} \end{aligned}$$
From Eqs. (i) and (ii),
$\quad \frac{f}{\pi r^2} \times \frac{L}{l}=\frac{\mathrm{f}}{\pi r^2} \times \frac{L}{l^{\prime}}$
$\therefore$ $l=l^{\prime}$
$$\text { ( } \because \text { Both wires are of same material, hence, Young's modulus will be same.) }$$
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 $2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$ )
$$ \begin{aligned} \text { Length of steel cable } l & =9.1 \mathrm{~m} \\ \text { Radius } r & =5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m} \\ \text { Tension in the cable } F & =800 \mathrm{~N} \\ \text { Young's modulus for steel } Y & =2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \\ \text { Change in length } \Delta l & =? \\ \text { Young's modulus }(Y) & =\frac{F}{A} \times \frac{l}{\Delta l} \Rightarrow \Delta l=\frac{F}{\pi r^2} \times \frac{l}{Y} \\ & =\frac{800}{3.14 \times\left(5 \times 10^{-3}\right)^2} \times \frac{9.1}{2 \times 10^{11}} \\ & =4.64 \times 10^{-4} \mathrm{~m} \end{aligned} $$