Physics

The best conductor of electricity is Silver (Ag) at room temperature. It is not used in normal wiring even though being the best conductor because of its high cost.

• Electrical Conductivity ($\sigma$): $6.3 \times 10^ {7} \ \text {S/m}$

• Electrical Resistivity ($\rho$): $1.59 \times 10^ {-8} \ \Omega \cdot \text {m}$ (the lowest among all metals)

A -> m_A = 1 kg, P_A = P

B ->  m_B = 2 kg,   P_B = P

$?kE=\frac{P^2}{2m}$

$\therefore \frac{k{E}_A}{k{E}_B}=\frac{\left(\frac{P^2}{2mA}\right)}{\left(\frac{P^2}{2m_B}\right)}=\frac{m_B}{m_A}=\frac{2}{1}$

$7.2*\frac{5}{18}=2m/s$ [V_A = V_B = 2m/s]

$f{\text{'}}_B=f_A\left(\frac{v+v_B}{v-v_A}\right)$

$\downarrow$     

App freq heard by car B

$f{\text{'}}_B=676\left[\frac{340+2}{340-2}\right]=676*\frac{342}{338}=684Hz$  

Similarly, $f{\text{'}}_A=f_B\left(\frac{v+v_B}{v-v_B}\right)=684Hz$

$\therefore$ Beat frequency heard by both = 684 – 676 = 8Hz

radius r = 0.5 mm

$\sigma =5*10^{7}s/m$

$\overrightarrow{E}=10*10^{-3}V/m=10^{-2}V/m$

As we know

$J=\sigma E$

$J=\frac{i}{A}=\sigma E$

$i=\sigma EA=5*10^7*10^{-2}*\pi *{\left(0.5*10^{-3}\right)}^2=1.25\pi *10^{-1}A=125\pi mA=x^3\pi mA$

$$ $\therefore$  x = 5

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

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

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