Physics

$L\frac{di}{dt}=V?2di=3tdt?2{\displaystyle \underset{0}{\overset{i}{?}}di=3}{\displaystyle \underset{0}{\overset{4}{?}}tdt?i=12A}$

$U=\frac{1}{2}L{i}^2=\frac{1}{2}*2*12^2=144J$

Indeed, it is an easy chapter of Class 12 Physics. In this chapter, you will learn about the foundational concepts like Gauss's law and electric fields.

Let the charge of on each drop be q and radius of each drop be r.

  $\frac{kq}{r}=2$

When all drops are joined, radius,

$r\text{'}={\left(512\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}r=8r$

Potential of the new drop,

$V=\frac{k.512q}{8r}=64\frac{kq}{r}=128V$

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

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

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

$v=\frac{1}{\sqrt[]{\epsilon \mu }}=\frac{1}{\sqrt[]{2{\epsilon }_0.2{\mu }_0}}=\frac{1}{2\sqrt[]{{\epsilon }_0{\mu }_0}}=\frac{c}{2}=15*10^{7}m/s.$

$\frac{N{}_p}{N_s}=\frac{V_p}{V_s}\Rightarrow N_p=\frac{V_p}{V_s}*N_s=\frac{220}{12}*24=440$

Assume the length of train be I and its acceleration be a.

$v^2=u^2+2al\Rightarrow al=\frac{v^2-u^2}{2}$

Velocity when middle point crosses the post,

$V_m=\sqrt[]{u^2+2a\frac{\mathcal{l}}{2}}.=\sqrt[]{u^2+\frac{v^2-u^2}{2}}=\sqrt[]{\frac{u^2+v^2}{2}}$

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2*3.14}{\sqrt[]{6.67*10^{-11}*6*10^{24}}}\left[{\left(8*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$v=\frac{p}{m}\Rightarrow v_p:v_d:v_{\alpha }=1:\frac{1}{2}:\frac{1}{4}=4:2:1$

$F_{mag}=qvB\Rightarrow F_P:F_d:F_{\alpha }=1*4:1*2:2*1=2:1:1$