Class 12th

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$           

$={\left(Sinx\right)}_0^{\pi /2}=1$           

$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$           

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$

The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux. 

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

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

For JEE Main exam, the Electric Charges and Fields is an important chapter. It is a part of Electrostatics unit. The weightage of this unit is around 6-9% of the total Physics section. You should expect around 3-5 questions in this unit including the questions from the class 12 physics chapter 1.

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

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

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

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

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

Given $\left(x\right)=\{\begin{array}{l}2sin\left(\frac{-\pi x}{2}\right),x<-1\\ \left|a{x}^2+x+b\right|,-1\le x\le 1\\ sin\pi x,x>1\end{array}$

If f (x) is continuous for all $x\in R$ then it should be continuous at x = 1 & x = -1

At x = -1, L.H.L = R.H.L. Þ 2 = |a + b - 1|

->a + b – 3 = 0  OR  a + b + 1 = 0 . (i)

-> a + b + 1 = 0 . (ii)

(i) & (ii), a + b =-1

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