Class 12th

$p=\frac{h}{\lambda }$  

So, two photons having equal linear momenta have equal wavelength. As wavelength decreases, momentum and energy of photon increases.

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here $\begin{array}{cc}\hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill \end{array}\begin{array}{cc}\hfill \begin{array}{l}\\ \end{array}\hfill & \hfill \hfill \end{array}$

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right|\begin{array}{cc}\hfill \begin{array}{l}=\left(42-22\right)-2\left(14+11\right)-3\left(-4-6\right)\\ \end{array}\hfill & \hfill =20-50+30=0\hfill \end{array}$

For infinite solution ${}_{}{}_{}{}_{}$

20a – 8b – 4c = 0 5a = 2b + c

Given $\frac{dy}{dx}=\frac{x{y}^2+y}{x}=y^2+\frac{y}{x}$

OR $\frac{dy}{dx}-\frac{y}{x}=y^{2}OR\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{x}.\frac{1}{y}=1.........\left(i\right)$

=> $\frac{x}{y}=-\frac{x^2}{2}+c..........\left(ii\right)$      

Since curve intersect x + 2y = 4 at x = -2 then y = 3 so

From (ii) $\frac{-2}{3}=-2+cORc=2-\frac{2}{3}=\frac{4}{3}$

put x = 3, then $\frac{3}{y}=\frac{-9}{2}+\frac{4}{3}=\frac{-19}{6}$

$\Rightarrow y=\frac{-18}{19}$  

Hydrogen peroxide reduces iodine to iodide ion is basic medium as;

$H_2{O}_2+2O{H}^{-}+l_2\to O_2+2l^{-}+2H_{2}O$

Given f (k) = $\{\begin{array}{l}k+1,kisodd\\ k,kiseven\end{array}$  

$?g:A\to A$ such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ * 1 = 10⁵

Here A is CH₃ - CH₂ – CH₂ – COO – CH₂ – CH₃

The given sequence of reaction is:

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-OH\underset{}{\overset{Oxidation}{\to }}C{H}_3-C{H}_2-C{H}_2-COOH$

Ce, Pr, Nd, Tb and Dy are the only lanthanoids which shows + 4 O.S, so can form MO₂, but Yb does not shows +4 O.S so can't form MO₂.

Only 1° aromatic amines give stable diazonium salt on reaction with nitrous acid

Here 1° aromatic amine is  which gives most stable diazonium salt

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$