Class 12th

When electric field is parallel then it would provide zero flux.

Using magnetic field due to straight wire :

$B=\frac{{\mu }_{0}i}{4\pi r}\left(sin\alpha +sin\beta \right)$

$=\frac{10^{-7}*1.5}{\left(\frac{0.09}{2\sqrt[]{3}}\right)}*\left(sin60°+sin60°\right)=10^{-5}T$

So, magnetic field due to three wires

= 3 * 10^-5 T

inside the plane

${\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^{3/4}{\left(x+2\right)}^{5/4}}={\displaystyle \int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{\left(x+2\right)}^2}put\frac{x-1}{x+2}}=t\Rightarrow \frac{3}{{\left(x+2\right)}^2}dx=dt}$

$\therefore \frac{1}{3}{\displaystyle \int \frac{dt}{t^{3/4}}=\frac{1}{3}.\frac{t^{1/4}}{\frac{1}{4}}+C=\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{1/4}+C}$

Comparing E = 20 cos (2 * 10¹⁰ t – 200x) V/m to

$E=E_{0}cos\left(\omega t-kx\right)v/m$              

$\omega =2*10^{10},K=200$               

Speed = $\frac{2*10^{10}}{200}=10^{8}m/s$  

R.I. $=\frac{C}{speed}=\frac{3*10^8}{10^8}=3$  

$NowR.I.=\sqrt[]{{\epsilon }_r{\mu }_r}$

$3=\sqrt[]{{\epsilon }_r*1}$

${\epsilon }_r=9$              

Students should prepare complete syllabus when they have time to prepare. However, you can use the list of highweightage chapters in last minute revision for scoring well.

• The p-Block Elements: This chapter holds a high weightage of 8–10 marks. (in the latest syllabus this is deleted)

• Aldehydes, Ketones, and Carboxylic Acids: This chapter contributes around 8–10 marks.

• Biomolecules: This chapter accounts for around 8 marks.

• Chemical Kinetics: This chapter holds a high weightage of 5-6 marks. 

• The d- and f-Block Elements: This chapter contributes around 5-6 marks.

• Amines: This chapter contributes around 5-6 marks.

Capacitor makes potential difference constant.

Using faraday's law magnetic field should be outward and decreasing with time

$LHS=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to \infty }{lim}\frac{-1}{h}ln\left(\frac{1-\frac{h}{a}}{1+\frac{h}{b}}\right)=\underset{h\to 0}{lim}\frac{ln\left(1-\frac{h}{a}\right)}{a\left(\frac{-h}{a}\right)}+\underset{h\to 0}{lim}\frac{ln\left(1+\frac{h}{b}\right)}{b\left(\frac{h}{b}\right)}=\frac{1}{a}+\frac{1}{b}.............\left(i\right)$

$RHS=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{cos2x-1}{x^2+1-1}\left(\sqrt[]{x^2+1}+1\right)=\underset{x\to 0}{lim}\frac{-2si{n}^{x}x}{x^2}*2=-4...............\left(ii\right)$

$and\underset{x\to 0}{lim}f\left(x\right)=k.............\left(iii\right)$

$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=-4=k$

From (i), (ii) and (iii) we get $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$