Class 11th

Let x = correct answer, y = incorrect answer

  $\therefore 3x-2y=5,x+y\le 5,x,y\in w$              

 only possible (x, y) is (3, 2)

$\therefore$ Required number of ways = 

Eutrophication of water body results in loss of biodiversity.

$AgC\mathcal{l}+2N{H}_3\to \underset{\left(Solublecomplex\right)}{\left[Ag{\left(N{H}_3\right)}_2\right]C\mathcal{l}}$

Size of hydrated ion $\propto$ $\frac{1}{Ionicmobility}$

Size of hydrated ions $\propto \frac{1}{Ionicsize}$

Size of hydrated ions ; $B{e}^{+2}>M{g}^{+2}>C{a}^{+2}>S{r}^{+2}$

Ionic mobility ; $B{e}^{+2}<M{g}^{+2}<C{a}^{+2}<S{r}^{+2}$

  $\frac{sin\theta sin\theta }{cos\theta }+\frac{sin\theta }{cos\theta }=2sin\theta cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }\left[sin\theta +1\right]=2sin\theta cos\theta$            

$\therefore$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$  

$\theta =\frac{3\pi }{2},\theta =\frac{\pi }{6},\frac{5\pi }{6}$           

$\theta =\frac{3\pi }{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{\pi }{3}+cos\frac{5\pi }{3}=4$  

$\therefore$ T + n(s) = 4 + 5 = 9

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

$\frac{x^2-5x+6}{x^2-9}\ge -1\&\frac{x^2-5x+6}{x^2-9}\le 1$

$\frac{2x+1}{x+3}\ge 0,x\ne -3\&\frac{1}{x+3}\ge 0,x\ne -3\Rightarrow x>-3$

$x\in \left[-\frac{1}{2},\infty \right]...........(i)$

$x^2-3x+2>0and{x}^2-3x+1\ne 0$

(x – 2) (x – 1) > 0 and $x\ne \frac{3\pm \sqrt[]{5}}{2}$

$x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

From (i) and (ii)   $x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

Number of electrons in

$P{H}_3=15+3=18$

$B_2{H}_6=5*2+6=16$

$\begin{array}{l}CC{l}_4=6+17*4=6+68=74\\ N{H}_3=7+1*3=10\\ LiH=3+1=4\\ BC{l}_3=5+17*3=56\end{array}$

$B_2{H}_6\&BC{l}_3$ are e^- deficient molecules. B₂H₆ is dimer of BH₃, both compound has 6e^- only.