Class 11th

Range R = $\frac{u^{2}sin2\theta }{g}$

For 42° and 48° Range will be same

$H_{max}\alpha$  maximum q

So maximum height will be for 48°

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

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

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

${\epsilon }_r=9$              

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

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

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

De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6*10^{-34}}{\sqrt[]{3*9*10^{-31}*1.38*10^{-23}*300}}$

$\begin{array}{l}\lambda =6.26*10^{-9}m\\ \lambda =6.26nm\end{array}$                

For circular motion F_{net   $=\frac{M{V}^2}{R}$}

_{$\sqrt[]{2F}+F\text{'}=\frac{M{V}^2}{R}$}

$\sqrt[]{2}\frac{GMM}{{\left(\sqrt[]{2}R\right)}^2}+\frac{GMM}{{\left(2R\right)}^2}=\frac{M{V}^2}{R}$                

$\frac{GM}{R}\left[\frac{1}{\sqrt[]{2}}+\frac{1}{4}\right]=V^2$              

$V=\frac{1}{2}\sqrt[]{\frac{GM}{R}\left(2\sqrt[]{2}+1\right)}$

  $\sim \left(\sim p\vee q\right)=p\wedge \sim q$

For every a, there  must be a² – 2. So, there will be infinitely many pairs (a, b)

$S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}..........\infty$

$\frac{S}{5}=\frac{7}{5^2}+\frac{9}{5^2}+\frac{13}{5^4}+------\infty$

$Let\frac{4S-7}{5}=t$

$\frac{4t}{5}=\frac{2}{25}\left\{\frac{1}{1-\frac{1}{5}}\right\}=\frac{1}{10}$

$t=\frac{1}{8}$

$\frac{4S-7}{5}=\frac{1}{8}$

$S=\frac{61}{32}$

$\therefore 160S=305$

The parabola : ${\left(x-\frac{1}{2}\right)}^2=y-\frac{3}{4}$  

-> y = x² – x + 1 . (i)

$P\left(-\frac{1}{2},\frac{7}{4}\right)$                

$N_P:y=\frac{x}{2}+2.........\left(ii\right)$                

(i) & (ii) -> Q (2, 3)

$P{Q}^2=\frac{125}{16}$                

$a^{10}{b}^{10}{\left(\frac{1}{b}+\frac{2}{a}+4\right)}^{10}$

$a^{10}{b}^{10}\frac{10!{\left(\frac{1}{b}\right)}^{r_1}{\left(\frac{2}{a}\right)}^{r_2}.4^{10-r_1-r_2}}{r_1!r_2!\left(10-r_1-r_2\right)!}$

r₁ = 2, r₂ = 3

$a^7{b}^{8}is\frac{10!2^3.4^{10-2-3}}{2!3!5!}=\frac{2^{13}.10!}{2!3!5!}=2^{16}.315$    

k = 315