Physics

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

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

Let m_AB = m = 1 kg

AB = 0.4 m =  $\mathcal{l}$

d = OM = $\sqrt[]{0.16-0.04}=\sqrt[]{0.12}$

$d=2\sqrt[]{3}*10^{-1}$  

$I_{ABaboutO}=\frac{m{\mathcal{l}}^2}{12}+m{d}^2$           

$\therefore l_{hexagon,aboutO}=6\left[\frac{m{\mathcal{l}}^2}{12}+m{d}^2\right]=\frac{m{\mathcal{l}}^2}{2}+6m{d}^2$       

$=\frac{1*{\left(0.4\right)}^2}{2}+6*1*{\left(2\sqrt[]{3}*10^{-1}\right)}^2=\frac{0.16}{2}+6*4*3*10^{-2}$   

= 0.08 + 0.72 = 0.8 kgm² = 8 * 10^-1 kgm²

Flux through the 6 sides of square (i.e. cube)

${?}_T=\frac{q_{in}}{{\epsilon }_0}=\frac{12*10^{-6}C}{{\epsilon }_0}$

$\therefore$ Flux through a square

$?=\frac{{?}_T}{6}=\frac{12*10^{-6}}{{\epsilon }_0*6}=\frac{2*10^{-6}}{{\epsilon }_0}$  

$\Rightarrow ?=225.98*10^{3}N{m}^2/C?226N{m}^2/C$

$T=27°C,P=1atm=10^{5}N/m^2,V_{rms}=200m/s$

As we know

$v_{rms}=\sqrt[]{\frac{3RT}{M}}\Rightarrow \frac{v_{rms}}{v{\text{'}}_{rms}}=\frac{\sqrt[]{\frac{3RT}{M}}}{\sqrt[]{\frac{3RT\text{'}}{M}}}$

$V{\text{'}}_{rms}=200*\frac{2}{\sqrt[]{3}}=\frac{x}{\sqrt[]{3}}$

x = 400

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$=\frac{\sigma \pi a^2*a-\frac{\sigma \pi a^2}{4}*\frac{3a}{2}}{\sigma \pi a^2-\frac{\sigma \pi a^2}{4}}=\frac{\frac{5}{8}\sigma \pi a^3}{\frac{3}{4}\sigma \pi a^2}=\frac{5}{8}\sigma \pi a^2*\frac{4}{3}\sigma \pi a^2=\frac{5}{6}a$

Travelling wave is represented by $f\left(t\pm \frac{x}{v}\right)$ $\frac{}{}$

Thus, y = A sin (15x – 2t) represents a travelling wave

$\lambda =\frac{h}{p}or\lambda =\frac{h}{mv}$

${\lambda }_p={\lambda }_{\alpha }\left[Given\right]$

$\frac{h}{m_p{V}_p}=\frac{h}{m_{\alpha }{v}_{\alpha }}\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{m_{\alpha }}{m_p}$

$\Rightarrow \frac{v_p}{v_{\alpha }}=\frac{4m}{m}=\frac{4}{1}$ 0r 4: 1

Kindly consider the following figure

${\lambda }_{Red}>{\lambda }_{violet}$

$\beta =\frac{\lambda D}{d}\Rightarrow$ Fringe width

If the source of light used in a Young's double slit experiment is changed from red to violet : consecutive fringe lines will come closer.