Class 12th

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 $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

Shiksha's NCERT notes are extremely useful for efficient preparation. We offer structured chapter-wise notes for the latest CBSE  syllabus and provide concise summaries and key formulas. These notes are designed for quick and last-minute revision. These short revision notes offer step-by-step explanations for conceptual clarity and include important questions from previous years' papers and NCERT Textbooks. Our notes are equally beneficial for competitive exams like JEE Mains, NEET and other exams as well.

Energy of electron in first existed state will be -3.4 eV.

So total energy difference will be (2.6 + 3.4) eV.

Wavelength $\left(\lambda \right)=\frac{1242eV-nm}{6eV}=207nm$

$Frequency=\frac{c}{\lambda }=\frac{3*10^8}{207*10^{-9}}=1.45*10^{9}MHz$