Class 12th

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

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

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

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

$k\ne 0\Rightarrow k=4$

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

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

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

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

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

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

Yes, candidates can get admission in Darshan University with 60% in Class 12. The minimum required aggregate for the same is 50%. Candidates must submit the marksheet for Class 12 to apply for and confirm admission at the university in the BSc course. Admissions to this course are merit-based, and merit of Class 12 is accepted. 

$M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill c_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$    

$M^{T}M=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill b_1\hfill & \hfill c_1\hfill \\ \hfill a_2\hfill & \hfill b_2\hfill & \hfill c_2\hfill \\ \hfill a_3\hfill & \hfill b_3\hfill & \hfill c_3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right]$

$Tr\left(M^{T}M\right)=a_1^2+b_1^2+c_1^2+a_2^2+b_2^2+c_2^2+a_3^2+b_3^2+c_3^2=7$           

all   $a_i,b_i,c_i\in \left\{0,1,2\right\}for$ i = 1, 2, 3

Case 1 7 one's and two zeroes which can occur in ways

Case 2 One 2 three 1's five zeroes =

total such matrices = 504 + 36 = 540

 $E=\frac{\lambda }{4\pi {\epsilon }_{0}a}\left[sin\alpha +sin\alpha \right]=\frac{\lambda sin\alpha }{2\pi {\epsilon }_{0}a}$

$\Rightarrow E=\frac{Q}{L2\pi {\epsilon }_0*\left(\frac{\sqrt[]{3}L}{2}\right)}*\frac{1}{2}=\frac{Q}{2\sqrt[]{3}\pi {\epsilon }_0{L}^2}$

$tan\alpha =\frac{\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\sqrt[]{3}L/2}\Rightarrow \alpha =\frac{\pi }{6}\Rightarrow sin\alpha =\frac{1}{2}$

As height of image is less than height of image and has same orientation as that of object, so mirror must be convex.