Physics

$E=56.5sin\left(w\left(\raisebox{1ex}{$t-x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)\right)N/c$

$E_0=56.5$

${\epsilon }_0=8.85*10^{-12}$

C = 3 * 108

l = $\frac{1}{2}{\epsilon }_0{E}_0^{2}C$

$=\frac{1}{2}*8.85*10^{-12}*{\left(56.5\right)}^2*3*10^8$

$=42377.11*10^{-4}$

$l=4.24w{m}^{-2}$

$\eta =1-\frac{T_2}{T_1}=1-\frac{100}{400}=\frac{3}{4}=75\mathrm{\%}$

Purely inductive circuit

$\theta =\frac{\pi }{2}$

$cos\frac{\pi }{2}=0$

Average power = 0

  $\frac{K_2}{k_1}=9\frac{A_1}{A_2}=2,\frac{L_1}{L_2}=2$

  $\frac{450-T}{\frac{L_1}{A_1{k}_1}}=\frac{T}{\frac{L_2}{k_2{A}_2}}\Rightarrow \left(450-T\right)\frac{L_2}{K_2{A}_2}=\left(T\right)\frac{L_1}{A_1{k}_1}$             

$\left(450-T\right)=T\left(\frac{L_1}{L_2}\frac{k_2{A}_2}{k_1{A}_1}\right)$                          

$\Rightarrow 450-T=T\left(2*9*\frac{1}{2}\right)=9T$               

10 T = 450

T = 45° C

for $\pi R^{2}Area\to \text{'}l\text{'}$ ${}^{}$ current

1 unit Area $\frac{l}{\pi R^2}$ $\frac{}{{}^{}}$

$\pi r^2\to \frac{l}{\pi R^2}*\pi r^2$

$i=\frac{l{r}^2}{R^2}$

Now, consider Amperian loop of radius small 'r' ln Amperian loop magnetic field will be tangential to the amperian loop.

$$ ${\displaystyle ?\overrightarrow{B}.d\overrightarrow{i}={\mu }_0{l}_{enclosed}}$   (Ampere circuital law)

$B=\frac{{\mu }_0}{2\pi }\frac{l}{R^2}r$

$B\propto r$

$\frac{E_A}{E_B}=\frac{-\frac{GM{m}_A}{2*3r}}{\frac{-GM{m}_B}{2*4r}}=\frac{m_A}{m_B}*\frac{4}{3}=\frac{4}{3}*\frac{4}{3}=\frac{16}{9}$

  $\rho =10^{3}kg/m^3$

g = 10 m/s²

Final height in both vessels

$=\frac{100+150}{2}=125cm$

So, less in U =    ${\displaystyle \int A*0.25*g*0.25}$

= $10^3*625*10^{-4}*10*16*10^{-4}$  

= 625 * 16 * 10^-4

= 1J.

(Independent of distance)

$E_1=E_2=\frac{\sigma }{2{\in }_0}$

  $a=\mu g=0.4*10=4m/s^2$

Slipping stops when V_bag = V_conveyar

-> 0 + at = 2

-> t =    $\frac{2}{a}=\frac{2}{4}=.5sec$

So, x =    $\frac{1}{2}a{t}^2=\frac{1}{2}*4*\frac{1}{4}=0.5m$

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$