Class 12th

$h{f}_1=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R*\frac{8}{9}$

$h{f}_2=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R*\frac{3}{4}$

$\frac{f_1}{f_2}=\frac{8/9}{3/4}=\frac{32}{27}$

f2 = f1 * $\left(\frac{27}{32}\right)$

$=\left(2.92*10^{15}\right)*\left(\frac{27}{32}\right)$

$=2.46*10^{15}Hz$

Kindly consider the following figure

For mirror

Mean $=np$ , variance $=npq$

$np-npq=1;p+q=1$

$n^2{p}^2-n^2{p}^2{q}^2=11$

We get   $q=\frac{5}{6},p=\frac{1}{6},n=36\mathrm{}$  Probability of ' 3 ' success $={}^{36}{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{33}$

$\bar{a}+\bar{b}+\bar{c}+\bar{d}=0$

Magnitude of all vectors will be same as well as angle between these vectors will also be same. $\bar{a}+\bar{b}+\bar{c}=-\bar{d}$

$\begin{array}{c}|\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{b}+2\bar{b}\cdot \bar{c}+2\bar{c}\cdot \bar{a}=|d{|}^2\\ \mathrm{}1+1+1+2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =1\\ 6\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-2\\ \mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-\frac{1}{3}\\ \alpha ={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\left(\frac{-1}{3}\right)=\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\frac{1}{3}\end{array}$

In the integral J, substitute x + 1 = t

$\Rightarrow dx=dtand{x}^2+2x=\left(t^2-1\right)$            

Now   $J={\displaystyle \underset{1}{\overset{e}{\int }}\frac{e^{\frac{t^2-2}{2}}}{t}}dtandK={\displaystyle \underset{1}{\overset{e}{\int }}t}lnt{e}^{\frac{t^2-2}{2}}dt$

Hence   $\left(J+K\right)={\displaystyle \underset{1}{\overset{e}{\int }}{e}^{\frac{t^2-2}{2}}}\left(\frac{1}{t}+tlnt\right)dt$

$={\left(e^{\frac{t^2-2}{2}}lnt\right)}_{t=1}^{t=e}=e^{\frac{e^2-2}{2}}={\left(\sqrt[]{e}\right)}^{e^2-2}$

TSA  of cube $=6a^2$

$\frac{d}{dt}\left(6a^2\right)=4.8;12a\frac{da}{dt}=4.8;a\frac{da}{dt}=0.4;\frac{dv}{dt}=\frac{d}{dt}\left(a^3\right)\Rightarrow 3a\left(a\frac{da}{dt}\right)$

$3*15*0.4=3*15*\frac{04}{10}\Rightarrow 18$

Intersity a number of photons kinetic Energy a f

P(W) = $\frac{1}{3};$  P(B) =   $\frac{2}{3}$

$\Rightarrow p=\frac{1}{3};q=\frac{2}{3}$           and r = 4 or 5 and n = 5

Use   $P\left(r\right){=}^n{C}_r{p}^r{q}^{n-r}$

 P(4) + P(5)

${=}^5{C}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^1{+}^5{C}_5{\left(\frac{1}{3}\right)}^5$            

$=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}=\frac{11}{243}$  

$R=\frac{\sqrt[]{2Em}}{B.q}$

$\Rightarrow R*\frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_1}{R_2}=\frac{\sqrt[]{4}}{2}*\frac{3}{\sqrt[]{16}}=\frac{3}{4}$

$\Rightarrow R_2=\frac{4R_1}{3}$

$sin\theta =\frac{d}{R}\Rightarrow \theta \alpha \frac{1}{2}\Rightarrow {\theta }_2<{\theta }_1$