Physics Mechanical Properties of Solids

Stress = Y * strain

=> $\frac{T_1}{A}=Y*-\frac{\left({\mathcal{l}}_1-\mathcal{l}\right)}{\mathcal{l}}\left(i\right)$

$\frac{T_2}{A}=Y*\frac{\left({\mathcal{l}}_2-\mathcal{l}\right)}{\mathcal{l}}\left(ii\right)$

=> $\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}$

$\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_2}{T_1-T_2}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$            

As we know that

$Y=\frac{FL}{A?L}$          

$?L=0.04m=\frac{FL}{AY}...............\left(i\right)$           

If length and diameter both are doubled

$?L\text{'}=\frac{F.2L}{4A.Y}=\frac{FL}{2Y}=0.02m=2cm$       

Energy density (u) = $\frac{1}{2}*\sigma *\epsilon =\frac{1}{2}*Y\epsilon *\epsilon =\frac{1}{2}{\epsilon }^{2}Y$ $\frac{}{}\frac{}{}\frac{}{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}{\epsilon }^{2}YA$

Strain $\left(\epsilon \right)=\frac{\Delta \mathcal{l}}{{\mathcal{l}}_0}=\alpha \Delta T=10^{-5}*10=10^{-4}$ $\frac{}{{}_{}}{}^{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}*10^{-8}*10^{11}*10^{-2}=5$

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T₀ ( developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_1}{T_1-T_2}$

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T0 (developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ $$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

$E=150*{\left(0.5\right)}^2=\frac{150}{4}$

$fluxflowing?=EA=\frac{150}{4}*{\left(0.5\right)}^2$              

$=\frac{150}{16}$             

Gausses law $?=\frac{q}{{\epsilon }_0}$  

  $\frac{150}{16}=\frac{q}{{\epsilon }_0}$            

$=8.3*10^{-11}C$              

$Y=\frac{Mg{L}^3}{4b{d}^3\delta }$

For significant error in Y

$=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta }{\delta }$              

$=\frac{1*10^{-3}}{2}+\frac{3*10^{-3}}{1}+\frac{10^{-2}}{4}+3*\frac{0.01*10^{-1}}{0.4}+\frac{10^{-2}}{5}$             

= 0.0155

$Y=\frac{Mg{L}^3}{4b{d}^3?}$              

For significant error in Y

$=\frac{?M}{M}+\frac{3?L}{L}+\frac{?b}{b}+3\frac{?d}{d}+\frac{??}{?}$           

$=\frac{1*10^{?3}}{2}+\frac{3*10^{?3}}{1}+\frac{10^{?2}}{4}+3*\frac{0.01*10^{?1}}{0.4}+\frac{10^{?2}}{5}$              

= 0.0155

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young's modulus :  $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125*10^3}{2.355*2*10^{11}}=2.65*10^{-7}$

W = 10 kg m/s²

A = 100 cm²

^{$\mathcal{l}=20cm$}

Y = 2 * 10¹¹ N/m²

^{$Y=\frac{F\mathcal{l}}{A\Delta \mathcal{l}}\Rightarrow \Delta \mathcal{l}=\frac{F\mathcal{l}}{AY}$}

For elemental mass of length, dx

Change in its length

$\Delta \mathcal{l}=5*10^{-10}m$