Class 12th

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

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

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

C : 4x² + 4y² – 12x + 8y + k = 0

$?\left(1,-\frac{1}{3}\right)$               

Lies on or inside the C then

$4+\frac{4}{9}-12-\frac{8}{3}+k\le 0$

$\Rightarrow k\le \frac{92}{9}$               

Now, circle lies in 4^th quadrant centre

$\equiv \left(\frac{3}{2},-1\right)$               

$\therefore r<1\Rightarrow \sqrt[]{\frac{9}{4}+1-\frac{k}{4}}<1$               

$\Rightarrow \frac{13}{4}-\frac{k}{4}<1$

$\Rightarrow \frac{k}{4}>\frac{9}{4}$

$\Rightarrow k>9$

$\therefore K\in \left(9,\frac{92}{9}\right)$      

Given vertex is  (5, 4) and directrix 3x + y – 29 = 0

Let foot of perpendicular of (5, 4) on directrix be (x₁, y₁)

$\frac{x_1-5}{3}=\frac{y_1-4}{1}=\frac{-\left(-10\right)}{10}$               

$\therefore \left(x_1,y_1\right)=\left(8,5\right)$               

So, focus of parabola will be S = (2, 3)

Let P(x, y) be any point on parabola, then

$\Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2=\frac{{\left(3x+y-29\right)}^2}{10}$               

$\Rightarrow x^2+9y^2-6xy+134x-2y-711=0$               

And given parabola

$x^2+a{y}^2+bxy+cx+dy+k=0$               

$\therefore a=9,b=-6,c=134,d=-2,k=-711$               

$\therefore a+b+c+d+k=576$    

$\left(ta{n}^{-1}y\right)-x)dy=\left(1+y^2\right)dx$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$

$l.F=e^{{\displaystyle \int \frac{1}{1+y^2}dy}}=e^{ta{n}^{-1}y}$

$x.e^{ta{n}^{-1}y}{\displaystyle \int \frac{e^{ta{n}^{-1}y}ta{n}^{-1}y}{1+y^2}dy}$

Let

$e^{ta{n}^{-1}y}=t$

$=x{e}^{ta{n}^{-1}y}=e^{ta{n}^{-1}y}y-e^{ta{n}^{-1}y}+c...(i)$

$?$ It passes through (1, 0) = c = 2

Now put y = tan 1, then

ex = e – e + 2

$\Rightarrow x=\frac{2}{6}$