Class 12th

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$ ……… (i)

$2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ ……… (ii)

$\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$

Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

$\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$

$\therefore {\left(5h-8k\right)}^2+5r^2=816$

Kindly consider the following figure

$1<\left|z-3+2i\right|<4$

$z=a+iba,b\in I$              

⇒ 1 < (a 3)² + (b + 2)² < 16

in equivalent to 1 < ² + β² < 16  $\alpha =a-3\in l$               

$\left(\pm 2,0\right)\left(\pm 3,0\right),\left(0,\pm 2\right)\left(0,\pm 3\right)$             $\beta =b+2\in l$

Total 40 such points are possible

$4\mathrm{L}\mathrm{i}+{\mathrm{O}}_2?2{\mathrm{L}\mathrm{i}}_2\mathrm{O}\mathrm{}$ oxide

$2\mathrm{N}\mathrm{a}+{\mathrm{O}}_2?{\mathrm{N}\mathrm{a}}_2{\mathrm{O}}_2$ peroxide

$\mathrm{K}+{\mathrm{O}}_2?{\mathrm{K}\mathrm{O}}_2\mathrm{}$ superoxide

${\left(2x^3+\frac{3}{x^k}\right)}^{12}$

gen term =

${=}^{12}{C}_r{2}^{12-r}.3^r.x^{36-3r-rk}$

For constant term

36 – 3r – rk = 0

$\Rightarrow k=\frac{36-3r}{r}$

for r = 1, 2, 4

¹²C_r2^12-r>2⁸

Possible values of k = 3, 1

 $2Mn{O}_4+16HCl\to 2MnC{l}_2+2KCl+8H_{2}O+C{l}_2$

$KMn{O}_4$ oxidizes HCl to Cl₂, so HCl is not used in permanganate titration.

$\mathrm{U}$ and $\mathrm{H}$ are temperature dependent

${\mathrm{C}}_{\mathrm{P},\mathrm{m}}-{\mathrm{C}}_{\mathrm{V},\mathrm{m}}=\mathrm{R}$  (for 1 mole of ideal gas)

$\mathrm{d}\mathrm{U}={\mathrm{C}}_{\mathrm{V}}\mathrm{d}\mathrm{t}$

 No. of electron = 62 – 2 = 60

$68^{E{r}^{+3}}$ No. of electron = 68 – 3 = 65

$70^{Y{b}^{+2}}$ No. of electron = 70 – 2 = 68

$71^{Lu+3}$ No. of electron = 71 – 3 = 68

$65^{T{b}^{+4}}$ No. of electron = 65 – 4 = 61

$63^{E{u}^{+2}}$ No. of electron = 63 – 2 = 61

$65^{T{b}^{+2}}$ No. of electron = 65 – 2 = 63

$69^{T{m}^{+4}}$ No. of electron = 69 – 4 = 65

1 < a₁ < a₂ ……18 < 77

77 = 1 + (20 – 1) . d If n numbers are in A.P.

76 = 19 * d ⇒ d = 4

⇒ a₁ = 5

$\begin{array}{l}\therefore a_1+a_2+....+a_{18}\\ =\frac{18}{2}\left[2*5+17*4\right]=702\end{array}$          

$I_2+10HN{O}_3\left(conc\right)\to 2Hl{O}_3+10N{O}_2+4H_{2}O$