Class 12th

Emf = $\frac{-d?}{dt}$ $\frac{}{}$

When magnet passes through the centre rogion of loop, no emf is induced and the polarity of emf will be negative while entering the loop.

dA = 2r dr

$dq=\sigma *2\pi rdr$

$\Rightarrow E={\displaystyle \underset{0}{\overset{R}{\int }}\frac{kdqz}{{\left(z^2+r^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt[]{R^2+z^2}}\right]$

$\frac{dy}{dx}+\frac{e^y-2x}{2x^2}=0$  

$e^{-y}\frac{dy}{dx}-\frac{e^{-y}}{x}=-\frac{-1}{2x^2}$              

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

 I. F. $=e^{{\displaystyle \int \frac{dx}{x}}}=e^{lnx}=x$  

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$              

Biuret test is given by proteins and dipeptide & compounds having following groups

Hence positive test given by tripeptide & biuret.

Monomer of novolac is

Molecular mass of monomer unit

              = 12 * 7 + 1 * 8 + 16 * 2

              = 84 + 8 + 32

              = 124 g

Molecular mass of polymer = 963 g

$\therefore$ n * mass of monomer = mass of polymer + (n – 1) * 18

(n – 1) unit of H₂O is removed during formation

$\therefore 124*n=963+18\left(n-1\right)$

106n = 945

 $\therefore n=\frac{945}{106}=8.91\approx 9$

$n\left(n+1\right)x^2+2\left(2n+1\right)x+4$

$=n\left(n+1\right)x^2+\left\{\left(2n+2\right)+2n\right\}x+4$

${\displaystyle \sum _{n=1}^9\frac{x}{\left(nx+2\right)\left(\left(n+1\right)x+2\right)}}$

$={\displaystyle \sum _{n=1}^9\left[\frac{1}{nx+2}-\frac{1}{\left(n+1\right)x+2}\right]}$

$\left[\left(\frac{1}{x+2}-\frac{1}{2x+2}\right)+\left(\frac{1}{2x+2}-\frac{1}{3x+2}\right)+....+\left(\frac{1}{9x+2}-\frac{1}{10x+2}\right)\right]$

$\therefore \underset{x\to 2}{Lt}\frac{9x}{\left(10x+2\right)\left(x+2\right)}=\frac{9}{44}$              

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\frac{1+si{n}^{2}x}{1+{\pi }^{sinx}}+\frac{1+si{n}^{2}x}{1+{\pi }^{-sinx}}\right]dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx=\frac{\pi }{2}+\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-cos2x\right)dx}}$

$=\frac{\pi }{2}+\frac{1}{2}{\left[x-\frac{sin2x}{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

FeCl₃. 3H₂O i.e. [Fe (H₂O)₃], K₃ [Fe (CN]₆ and [Co (NH₃)₆]Cl₃ the last two complex are inner-orbital complex due to presence of strong ligand.

$\therefore {\Delta }_{0}of{K}_3\left[Fe{\left(CN\right)}_6\right]>{\Delta }_{0}of\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$

Because CN^- is strong ligand

$\therefore {\Delta }_0\propto \frac{1}{\lambda }$

More ${\Delta }_0$ smaller value of absorbed $\lambda$

$K_3\left[Fe{\left(CN\right)}_6\right]\to F{e}^{+3}=4s^{0}3d^{5}4p^0$

$\therefore u.e=1$

$\therefore \mu =\sqrt[]{1*\left(1+2\right)}B.M=\sqrt[]{3}B.M$

= 1.732 B.M

$\approx$ 2

$A^2=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^3=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^{2025}-A^{2020}$

$=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2024\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)-\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2019\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=A^6-A$

Number of replaceable H is 2.