Class 12th

$\alpha =sin36°=x\left(say\right)$

$\therefore x=\frac{\sqrt[]{10-2\sqrt[]{5}}}{4}$

$\Rightarrow 16x^2=10-2\sqrt[]{5}$

$\Rightarrow 16x^4-80x^2+20=0$

$\therefore 4x^4-20x^2+5=0$

$cot\left({\displaystyle {\sum }_{n=1}^{50}ta{n}^{-1}}\left(\frac{1}{1+n+n^2}\right)\right)$

$=cot\left(ta{n}^{-1}51-ta{n}^{-1}1\right)$

$=cot\left(co{t}^{-1}\left(\frac{52}{50}\right)\right)$

$=\frac{26}{25}$

The required probability

$=\frac{AreaofRegionPQCAP}{AreaofRegionABCA}$

$=\frac{\frac{1}{2}*8*6-\frac{1}{2}*2*4}{\frac{1}{2}*8*6}$

$=\frac{5}{6}$

Sol. first order reaction

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{{\mathrm{a}}_0-\mathrm{x}}$

$\mathrm{K}=\frac{2.303}{90}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{{\mathrm{a}}_0}{0.25{\mathrm{a}}_0}$

$=0.0154$

$\begin{array}{rr}& t=60\mathrm{\%}=\frac{2.303}{K}\mathrm{l}\mathrm{o}\mathrm{g}?\frac{a_0}{a_0}\dots .\left(2\right)\\ & =\frac{2.303}{0.0154}*(1-0.602)=59.51\mathrm{m}\mathrm{i}\mathrm{n}s\approx 60\end{array}$

Sol. 

$P_T=X_A\left(P_A^0-P_B^0\right)+P_B^0$

ATQ

$550=\frac{1}{4}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+4{\mathrm{P}}_{\mathrm{B}}^0$

$560=\frac{1}{5}\left({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0\right)+{\mathrm{P}}_{\mathrm{B}}^0$

$2200={\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0+5{\mathrm{P}}_{\mathrm{B}}^0$

${\mathrm{P}}_{\mathrm{A}}^0+3{\mathrm{P}}_{\mathrm{B}}^0=2200$

$\frac{{\mathrm{P}}_{\mathrm{A}}^0\pm 4{\mathrm{P}}_{\mathrm{B}}^0=2800}{{\mathrm{P}}_{\mathrm{B}}^0=600}$

${\mathrm{P}}_{\mathrm{A}}^0=400\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$

Sol. $Z=101$ belong to actinoids

104 belong to group 4

Kindly go through the solution

Be the vectors along the diagonals of a parallelogram having are 2.

$\therefore \frac{1}{2}\left|\overrightarrow{a}*\overrightarrow{b}\right|=2\sqrt[]{2}$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta =4\sqrt[]{2}$

$\Rightarrow \left|\overrightarrow{b}\right|sin\theta =4\sqrt[]{2}$ ……. (i)

And

$\Rightarrow \overrightarrow{c}.\overrightarrow{b}=-2{\left|\overrightarrow{b}\right|}^2=-128......(ii)$

$\Rightarrow \left|\overrightarrow{c}\right|=16\sqrt[]{2}.......(iii)$

From (ii) and (iii)

$\left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|cos\alpha =-128$

$\Rightarrow cos\alpha =\frac{1}{\sqrt[]{2}}$

$\alpha =\frac{3\pi }{4}$

$L_1:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$  

$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$               

Now,

$\overrightarrow{p}*\overrightarrow{q}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|=10\widehat{i}-8\widehat{j}-4\widehat{k}$               

and

  ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\widehat{i}-4\widehat{j}-4\widehat{k}$

  $\therefore S.D=\left|\frac{60+32+16}{\sqrt[]{100+64+16}}\right|=\frac{108}{\sqrt[]{180}}=\frac{18}{\sqrt[]{5}}$      

$L_1:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$  

$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$               

Now,

$\overrightarrow{p}*\overrightarrow{q}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|=10\widehat{i}-8\widehat{j}-4\widehat{k}$               

and

  ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\widehat{i}-4\widehat{j}-4\widehat{k}$

  $\therefore S.D=\left|\frac{60+32+16}{\sqrt[]{100+64+16}}\right|=\frac{108}{\sqrt[]{180}}=\frac{18}{\sqrt[]{5}}$