Class 11th

  $\begin{array}{l}2x+y=4\\ 2x+6y=14\end{array}\}y=2,x=3$              

B (1, 2)

Let C (k, 4 – 2k)

Now AB² = AC²

->5k² – 24k + 19 = 0

$\alpha =\frac{6+1+\frac{10}{5}}{3}=\frac{18}{5}$    

Now 15 (a + b)

$15\left(\frac{17}{5}\right)=51$                

  $f\left(x\right)=x^4-4x+1=0$              

$f\text{'}\left(x\right)=4x^3-4$

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

$\Rightarrow$ Two solution

  $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$                

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$              

$a{x}^2+a{y}^2+2ax-2ay=k$            

$\Rightarrow x^2+y^2+2x-2y=\lambda$              

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$

= 13 – 5

= 8

  $x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$               

Now,

a, b, c -> AP

1 – a, 1 – b, 1 – c -> AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z -> HP

  $\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x² – y² + 2xiy)

Case-I

x = 0

-y² = -y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$  

Mol wt A₂B = Mol wt of AB₃

 Hence (2A + B) = (A + 3B)

  $\therefore A=2B$

Atomic wt of A is two times of atomic wt of B.

$2NOC{l}_{\left(g\right)}?2NO\left(g\right)+C{l}_2\left(g\right)$

t = 0       2 mol/Lit             0            0

t_eq          2 – x                    x            x/2

$K_C=\frac{{\left[NO\right]}^2*\left[C{l}_2\right]}{{\left[NOCl\right]}^2}=\frac{x^2*x}{{\left(2-x\right)}^2*2}=\frac{x^3}{{\left(2-x\right)}^2*2}$      

$=\frac{0.4*0.4*0.4}{1.6*1.6*2}$ = 0.0125 = 125 * 10^-4

$\Delta V=2.4*10^{-26}m/sec$

And $\Delta x=10^{-7}m$  

According to Heisenberg's uncertainty

$\Delta v*\Delta x=\frac{h}{4\pi *m}$  

$\therefore m=\frac{h}{4\pi *\Delta v*\Delta x}=\frac{6.626*10^{-34}}{4*3.14*2.4*10^{-26}*10^{-7}}$

$\therefore$  m = 0.02198 Kg

Or 21.98 gm

Or 22 gm in the nearest integer.

Moles of $\mathrm{H}\mathrm{C}\mathrm{l}=$ moles of ${\mathrm{N}\mathrm{H}}_3$

$=2*$ moles of urea $=2*\frac{0.6}{60}=0.02$

$\Rightarrow \mathrm{N}.\mathrm{V}=0.02$

Kindly go through the solution

Conceptual.