Physics Mechanical Properties of Solids

$P=-B\left(\frac{\Delta v}{v}\right)$

or  $B=\frac{-P}{\left(\frac{\Delta v}{v}\right)}$

$B=\frac{-\rho gh}{\left(\frac{-0.005V_0}{V_0}\right)}\Rightarrow h=\frac{908*10^8*0.005}{10^3*9.8}$

$\Rightarrow h=500m$

For $W_1,$  

$\frac{\left(m+10+20\right)*10}{8*10^{-7}}\le 1.25*10^9$              

$\Rightarrow m\le 70kg$              

For W₂,

$\frac{\left(m+10\right)*10}{4*10^{-7}}\le 1.25*10^9$              

-> $m\le 40kg$               

Thus, $m\le 40kg$  

Tension in wire, $T=\frac{2*3*5*10}{3+5}=\frac{75}{2}N$

$\frac{T}{\pi r_{min}^2}=\frac{24}{\pi }*10^2$

$\Rightarrow r_{min}^2=\frac{75}{2*24}*10^{-2}=\frac{25}{16}*10^{-2}$

$\Rightarrow r_{min}=\frac{5}{4}*10^{-1}m=12.5cm$

Elastic deformation leads to the temporary change in a material's shape that is also reversible. It occurs for energy storage in devices, such as springs. It's also essential for understanding the flexibility of structural components like beams in bridges. This principle is used in engineering design. You can think vehicle suspension systems and even the resilience of buildings against wind forces.

When we talk about elastic materials, we're referring to those that really show off their elasticity. Think of natural rubber, synthetic polymers, including spandex (also known as Lycra) and nylon, and even metals such as spring steel when they're within their elastic limits. These materials are quite resilient. They can store potential energy and bounce back to their original shape after being stretched or compressed. This behaviour is determined by the Hooke's Law.

Elastic energy is stored in objects that can deform. They have several applications, including springs in vehicle suspension systems and wind-up clocks. This potential energy, present in items like stretched rubber bands, trampolines, and an archer's bow, is converted to kinetic energy upon release.

m = 10g                                                                       

$\mathcal{l}=50cm$

A = 2mm²

Y = 1.2 * 10¹¹N/m²

$\Delta x=x*10^{-5}m$  

As, $\frac{T}{A}=Y\frac{\Delta x}{\mathcal{l}}$  

$\Delta x=\frac{T\mathcal{l}}{AY}$

$T\mathcal{l}=V^{2}m$

=  $\frac{V^{2}m}{AY}$  

$=\frac{3600*10*10^{-3}}{2*10^{-6}*1.2*10^{11}}=\frac{1800*10^{-3}*10^6}{1.2}$

$=\frac{18}{12}*10^{-4}=\frac{3}{2}*10^{-4}=15*10^{-5}m$

So, x = 15

Slope = $\frac{\Delta L/W}{L}=\frac{\Delta L/L}{W}=\frac{1}{YA}$ $\frac{}{}\frac{}{}\frac{}{}$

$Y=\frac{1}{\left(Slope\right)A}$

$Y=\frac{1}{\left(2*10^{-6}\right)\left(0.25*10^{-5}\right)}$

$Y=2*10^{11}N/m^2$

  $\mathcal{l}=0.5m$

A ${}^{}{}^{}$ $=10^{-4}{m}^2$

Breaking stress, $\frac{F}{A}=5*10^{8}N/m^2$ $\frac{}{}{}^{}{}^{}$

$F=5*10^8*10^{-4}N$

F = 5 * 104N

T = $\frac{m{V}^2}{\mathcal{l}}\Rightarrow V_{\left(max\right)}^2==\frac{T\mathcal{l}}{m}$ $\frac{{}^{}}{}{}_{}^{}\frac{}{}$

$V_m^2=2.5*10^3$

V_max = 50m/s

Shear modulus = 25 * 109 N/m2

$\frac{F}{A}=\tau \frac{\Delta x}{h}\Rightarrow \Delta x=\frac{Fh}{A\tau }=\frac{18*10^4*0.15}{3600*10^{-4}*25*10^9}$

$=\frac{15*10^2}{200*25*10^5}$

$=\frac{3*10^2}{5*200}*10^{-5}$

$=\frac{60}{200}*10^{-5}=3\mu m$