Class 12th

Lyman Series n_f = 1, ni = 2, 3.

Paschen series n_f = 3, ni = 4, 5, 6 -

$\frac{1}{{\lambda }_1}=R{Z}^2\left(\frac{1}{1^2}-\frac{1}{4^2}\right)$

$\frac{1}{{\lambda }_2}=R{Z}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$

$\frac{1/{\lambda }_1}{1/{\lambda }_2}=\frac{1-\frac{1}{16}}{\frac{1}{9}-\frac{1}{16}}\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{15}{16}}{\frac{7}{9*16}}=\frac{15*9}{7}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\frac{7}{135}$

$l=l_{1}sin\omega t+l_{2}cos\omega t=\sqrt[]{l_1^2+l_2^2}\left[\left(\frac{l_1}{\sqrt[]{l_1^2+l_2^2}}\right)sin\omega t+\frac{l_2}{\sqrt[]{l_1^2+l_2^2}}cos\omega t\right]$

$=\sqrt[]{l_1^2+l_2^2}sin\left(\omega t+\alpha \right)=l_{0}sin(\omega t+\alpha$

$l_{rms}=\frac{l_0}{\sqrt[]{2}}=\frac{\sqrt[]{l_1^2+l_2^2}}{\sqrt[]{2}}$

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

[as probability of getting odd number = probability of getting even number = $\frac{1}{2}$ ]

Probability of getting an odd number for odd number of times =

  ${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$           

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$           

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$            

  $PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$         

$\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}.....\left(i\right)$

$\overrightarrow{a}.\overrightarrow{c}=7\overrightarrow{b}.\overrightarrow{c}=0$           

$\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{3}$          

$\overrightarrow{b}=2\widehat{i}+\widehat{k}\Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{5}\overrightarrow{a}.\overrightarrow{b}=-2+1=-1$           

From   $\left(i\right)\overrightarrow{a}.\overrightarrow{c}=\alpha {\left|\overrightarrow{a}\right|}^2-\beta$

$3\alpha -\beta =7..........\left(ii\right)$        

$\overline{b}.\overline{c}=\alpha \overline{b}.\overline{a}+\beta {\left|\overrightarrow{b}\right|}^2\Rightarrow -\alpha +5\beta =0.......\left(iii\right)$     

Solving $\alpha =\frac{5}{2}and\beta =\frac{1}{2}$  

$\Rightarrow 2{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=75$       

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) = $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

->b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$      

taking θ₁ > θ₂

Heat current = $\frac{{\theta }_1-\theta }{R_1}=\frac{\theta -{\theta }_2}{R_2}\Rightarrow \frac{{\theta }_1}{R_1}+\frac{{\theta }_2}{R_2}=\theta \left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

$\Rightarrow \theta =\frac{{\theta }_1{R}_2+{\theta }_2{R}_1}{R_1+R_2}$

$\lambda =\frac{hc}{E}=\frac{6.63*10^{-34}*3*10^8}{1.9*1.6*10^{-19}}$ = 654 nm (red)

According question, we can write

$\frac{C_1*C_2}{C_1+C_2}=\frac{15}{4}\left(C_1+C_2\right)\Rightarrow 4C_1{C}_2=15\left(C_1^2+C_2^2+2C_1{C}_2\right)$

$\Rightarrow 4\left(\frac{C_2}{C_1}\right)=15{\left(\frac{C_2}{C_1}\right)}^2+30\left(\frac{C_2}{C_1}\right)+15\Rightarrow 15x^2+26x+15=0,$ where x = $\frac{C_2}{C_1}$

$\Rightarrow x=\frac{-26\pm \sqrt[]{676-900}}{30}$

Since x cannot be real, so this Question has been dropped by NTA