Physics

f = 3GHz = 3 * 10⁹ Hz

${\lambda }_{vacuum}=\frac{v}{f}=\frac{3*10^{8}m/s}{3*10^9/s}=10^{-1}m=0.1m$

${\lambda }_{vacuum}=\frac{{\lambda }_{vacuum}}{{\mu }_{medium}}=\frac{0.1}{{\mu }_{medium}}$

${\mu }_{medium}=\sqrt[]{{\mu }_r{\in }_r}=\sqrt[]{1*2.25}=1.5$

$\therefore {\mu }_{medium}=\frac{0.1m}{1.5}=0.0667m=6.67cm=667*10^{-2}cm$

Sound level is given in dB = 10 log $\frac{{}_{}}{{}_{}}\frac{}{{}_{}}$ $\left(\frac{P_0}{P_i}\right)or10log\left(\frac{l}{l_0}\right)$

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 * 5 = 100 dB.

$\Delta \beta ={\beta }_2-{\beta }_1=10log\left(\frac{l_2}{l_1}\right)$

$-100=10log\left(\frac{l_2}{l_1}\right)$

$10^{-10}=\frac{l_2}{l_1}\Rightarrow l_2=10^{-10}{l}_1$

$P_2=10^{-10}{P}_1$

$P_2=10^{-10}*\left(0.1*10^3\right)\Rightarrow P_2=10^{-8}W=10^{-x}W$

$$ $\therefore$  x = 8

${\omega }_0=10^{5}rad/sec$

P = 16w, 120v at resonance

$P=\frac{v^2}{R}\Rightarrow 16=\frac{{\left(120\right)}^2}{R}\Rightarrow R=\frac{14400}{16}=900\Omega$

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