Physics

$B_0=\frac{E_0}{c}=\frac{800}{3*10^8}T$              

$F_{max}=ev{B}_0=1.6*10^{-19}*3*10^7*\frac{800}{3*10^8}=12.8*10^{-18}N$

$l_{CD}=\frac{1}{2}M{r}^2+M{\left(\frac{L}{2}\right)}^2$

$\Rightarrow 2.7=M\left(\frac{0.2^2}{2}+\frac{0.8^2}{4}\right)$              

M = 15 kg

Density, $\rho =\frac{M}{\pi r^{2}L}=\frac{15}{3.14*0.2^2*0.8}=1.49*10^{2}kg/m^3$  

Horizontal component of velocity of bomb equals to the velocity of plane. So, bomb and plane are in same vertical line always.

$V_A=\frac{1}{4\pi {\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt[]{a^2+s^2}}\right]$

$V_B=\frac{-1}{4\pi {\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt[]{a^2+s^2}}\right]$

$V_A-V_B=\frac{1}{2\pi {\epsilon }_0}\left[\frac{1}{a}-\frac{1}{\sqrt[]{a^2+s^2}}\right]$

$\frac{d{N}_B}{dt}=\lambda N_A-\lambda N_B=2\lambda N_B\left(0\right).e^{-\lambda t}-\lambda N_B$                

$\Rightarrow \frac{d{N}_B}{dt}+\lambda N_B=2\lambda N_B\left(0\right)e^{-\lambda t}$              

Solving it, $N_B\left(t\right)=N_B\left(0\right)\left(1+2\lambda t\right)e^{-\lambda t}$  

N_B (t) is maximum when,

$\frac{d{N}_B}{dt}=0\Rightarrow t=\frac{1}{2\lambda }$              

  $Tsin\theta =\frac{m{v}^2}{r}$

T cosθ = mg

$tan\theta =\frac{v^2}{rg}$              

$\Rightarrow v=\sqrt[]{rg}$              

Gauss law for magnetism states that total magnetic flux through a closed surface S will be zero. It is one of the four maxwell's equations that is useful in understanding the prinicples of electromagnetism. When understood in depth, the law implies that magnetic field lines form closed loops since they are continous. In short, it confirms that magnetic monopoles cannot exist.

Let, current through inductor,

$i_L=l_{L}sin\left(\omega t+{?}_1\right)$                

$l_L=\frac{200\sqrt[]{2}}{\sqrt[]{50^2+{\left(100*0.50\right)}^2}}=4A$              

${?}_1=ta{n}^{-1}\left(\frac{-100*0.50}{50}\right)=-\frac{\pi }{4}$             

Let, current through capacitor,

As  ${?}_2-{?}_1=\frac{\pi }{2}$  

$l=\sqrt[]{{\left(I_L\right)}^2+{\left(I_C\right)}^2}$

$=\sqrt[]{4^2+2^2}=4.47A$